How can an object with zero potential and kinetic energy ever move?












5












$begingroup$


I am not sure how to ask this question but I am learning about potential energy in my high school physics class. From the definition of potential energy, (energy stored in an object with the potential to convert into other type of energy), I don't understand how an object (let's say a ball) on the ground, which has zero kinetic energy and zero potential energy, can fall off a cliff and gain kinetic energy when it had no potential energy. How can it now be gaining energy? Also, another question that was already discussed in SE (but I didn't find my answer there), is why do we talk about the potential energy of a system (ball+Earth) but kinetic energy of an object (ball)?



Thanks.






newtonian-mechanics energy potential-energy







share|cite|improve this question











$endgroup$












  • $begingroup$
    Related: How does anything move? and links therein.
    $endgroup$
    – Qmechanic♦
    Jan 23 at 15:41


















5












$begingroup$


I am not sure how to ask this question but I am learning about potential energy in my high school physics class. From the definition of potential energy, (energy stored in an object with the potential to convert into other type of energy), I don't understand how an object (let's say a ball) on the ground, which has zero kinetic energy and zero potential energy, can fall off a cliff and gain kinetic energy when it had no potential energy. How can it now be gaining energy? Also, another question that was already discussed in SE (but I didn't find my answer there), is why do we talk about the potential energy of a system (ball+Earth) but kinetic energy of an object (ball)?



Thanks.






newtonian-mechanics energy potential-energy







share|cite|improve this question











$endgroup$












  • $begingroup$
    Related: How does anything move? and links therein.
    $endgroup$
    – Qmechanic♦
    Jan 23 at 15:41
















5












5








5


1



$begingroup$


I am not sure how to ask this question but I am learning about potential energy in my high school physics class. From the definition of potential energy, (energy stored in an object with the potential to convert into other type of energy), I don't understand how an object (let's say a ball) on the ground, which has zero kinetic energy and zero potential energy, can fall off a cliff and gain kinetic energy when it had no potential energy. How can it now be gaining energy? Also, another question that was already discussed in SE (but I didn't find my answer there), is why do we talk about the potential energy of a system (ball+Earth) but kinetic energy of an object (ball)?



Thanks.






newtonian-mechanics energy potential-energy







share|cite|improve this question











$endgroup$




I am not sure how to ask this question but I am learning about potential energy in my high school physics class. From the definition of potential energy, (energy stored in an object with the potential to convert into other type of energy), I don't understand how an object (let's say a ball) on the ground, which has zero kinetic energy and zero potential energy, can fall off a cliff and gain kinetic energy when it had no potential energy. How can it now be gaining energy? Also, another question that was already discussed in SE (but I didn't find my answer there), is why do we talk about the potential energy of a system (ball+Earth) but kinetic energy of an object (ball)?



Thanks.






newtonian-mechanics energy potential-energy




newtonian-mechanics energy potential-energy






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 23 at 20:46









Kevin

1053




1053










asked Jan 23 at 15:24









Lauren SinLauren Sin

436




436












  • $begingroup$
    Related: How does anything move? and links therein.
    $endgroup$
    – Qmechanic♦
    Jan 23 at 15:41




















  • $begingroup$
    Related: How does anything move? and links therein.
    $endgroup$
    – Qmechanic♦
    Jan 23 at 15:41


















$begingroup$
Related: How does anything move? and links therein.
$endgroup$
– Qmechanic♦
Jan 23 at 15:41






$begingroup$
Related: How does anything move? and links therein.
$endgroup$
– Qmechanic♦
Jan 23 at 15:41












6 Answers
6






active

oldest

votes


















6












$begingroup$

Where we define the potential energy to be $0$ in classical mechanics is arbitrary. All that matters is the change in potential energy. Since you are just learning this stuff I will assume you are in an algebra based physics class, so I will avoid using calculus here.



Potential energies are nice because they tell us how much work is done by a conservative force. More specifically, the work done by a conservative force is given by
$$W_{cons}=-Delta U$$
where $U$ is the potential energy associated with that conservative force. This is useful because we also know that the net work done on an object determines its change in kinetic energy
$$W_{net}=Delta K$$



So, if we consider your case where we just have one conservative force acting on the object, we can conclude that
$$Delta K=-Delta U$$



And so we see here that the only thing that determines how the motion of our object changes is just the change in potential energy. If we define the zero-point to be at the top of the cliff, then as the object falls its kinetic energy will grow and its potential energy will decrease and be negative when it hits the bottom of the cliff. If we define the zero-point to be at the bottom of the cliff then as the object falls its kinetic energy will grow and its potential energy will decrease to $0$ when it hits the bottom of the cliff. In either case the same thing happens because we have the same change in potential energy.




Also another question that was already discussed in SE (but i didn't find my answer there) is why do we talk about the potential energy of a system (ball+Earth) but kinetic energy of an object (ball)?




Typically in introductory physics classes we just consider the ball being in a uniform gravitational field and we don't even consider the Earth. However if you want to include the Earth in your system then when the ball falls the Earth will actually move slightly upwards to meet the ball. It will still be the case though that the work done by gravity on each object is related to the change in the potential energy of each object, which results in a change of kinetic energy.






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    "All that matters is the change in potential energy". That is true when calculating kinetic energy gained, but two balls at equal heights in the air, one above high ground an one above a cliff still have the same potential energy. Does a ball on top of a mountain have potential energy? Yes because the potential energy it accumulated while being lifted up there, even thou it is standing still.
    $endgroup$
    – Alex Doe
    Jan 23 at 17:51












  • $begingroup$
    @AlexDoe You can say that the potential energy increased as you lifted the ball to the top of the mountain, but if you set $h=0$ at the top then you can say the ball's potential energy is $0$. That is why it is better to say "the ball's potential energy increased"
    $endgroup$
    – Aaron Stevens
    Jan 23 at 18:40










  • $begingroup$
    I see your point. I think the opinion differences are in terminology. The way I was looking at it is like if I go to the mall and I set a limit as to how much I want to spend there (arbitrary). That doesn't mean that that's all the money (potential energy) that I have. Thanks for the input thou. Seems like I am a minority/alone on this side of the isle thou :)
    $endgroup$
    – Alex Doe
    Jan 23 at 19:10










  • $begingroup$
    @AlexDoe In your analogy it doesn't matter where you set your spending limit. You could spend more or less money than your limit and have "positive or negative" cash relative to that limit, but either way the new shirt you buy will change the amount of cash you have in your account by the same amount no matter what you decide your spending limit is.
    $endgroup$
    – Aaron Stevens
    Jan 23 at 19:14








  • 1




    $begingroup$
    @AlexDoe: I agree with you; I think it would be better to say "All that matters is differences in potential energy" (rather than specifically change). The "zero" is arbitrary, but you need to fix it once and stick with it for a whole calculation -- you can't just keep resetting your arbitrary zero and expect everything to work out.
    $endgroup$
    – ruakh
    Jan 23 at 20:04





















4












$begingroup$

The energy of an object consists of its internal energy (kinetic and potential energy possessed internally to an object at the atomic or molecular level) and external kinetic and potential energy. Its external kinetic and potential energy is always with respect to some external frame of reference.



The ball on the surface of the earth may have zero potential energy with respect to the surface of the earth, but has gravitational potential energy with respect to the center of the earth. It doesn’t move because the force of gravity downward on the ball is equal and opposite to the force exerted upward on it by the ground. But imagine if there was a trench that went from one side of the earth to the other. If the ball were positioned over the hole and allowed to fall, it would accelerate toward the center of the earth losing potential energy and gaining kinetic energy, the latter being a maximum at the center where its potential energy is zero. Then it would decelerate as it goes past the center as it gains potential energy and loses kinetic energy, ultimately stopping at the other side (ignoring air resistance).



Even the kinetic energy of the object is relative to the frame of reference where the velocity is measured. If you are in a car moving at velocity $v$ with respect to the road, it has zero kinetic energy in your frame of reference. It is not “moving” in your frame of reference, but is moving in the frame of reference of a person standing on the road. To that person the car has kinetic energy. The person standing on the road is “moving” with respect to your frame of reference in the car, so that person has kinetic energy with respect to you in the car. The person standing on the road has kinetic energy with respect to a frame of reference other than the earth (since the earth is rotating). And so on…



Bottom line, everything can be considered to have kinetic energy (be moving) with respect to some frame of reference and have potential energy (due to its position) with respect to some frame of reference.



Hope this help.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I like how you mention the arbitrariness of the "zero-point" of kinetic energy. I knew this was the case, but never realized how they are kind of the same idea. In a sense, just like how we define where $h=0$ for the zero-point of potential energy, we can also define what constitutes as $v=0$ for a sort of "zero-point" of kinetic energy.
    $endgroup$
    – Aaron Stevens
    Jan 23 at 16:56










  • $begingroup$
    Thanks. I like the way you call it "zero point" kinetic and potential energy.
    $endgroup$
    – Bob D
    Jan 23 at 17:02










  • $begingroup$
    Yeah it is probably not the most accurate terminology considering the use of "zero-point energy" in QM, but since we are in the classical realm for this question I think I will keep it haha
    $endgroup$
    – Aaron Stevens
    Jan 23 at 17:03






  • 2




    $begingroup$
    @Ed999 Bob never said anything contradicting what you just explained
    $endgroup$
    – Aaron Stevens
    Jan 23 at 18:44






  • 1




    $begingroup$
    @Ed999 I thought that's what I said.
    $endgroup$
    – Bob D
    Jan 23 at 19:18



















2












$begingroup$


I am not sure how to ask this question but I am learning about
potential energy (high school physics) and from the deffinition of a
potential energy (energy stored in an object with the potential to
convert into other type of energy)




this definition of potential energy, sounds more like the energy itself. More below




i don't understand how for eg. when we have an object (let's say a
ball) on the ground, it has zero kinetic energy and also zero
potential energy and now let's say the ball starts falling of a cliff
so it will be gaining kinetic energy but when it had no potential
energy, how can it be now gaining energy?




indeed, in the planet+object system, the true 0 potential, would be better put at the center of gravity the earth if the reference frame is attached to the earth, if the reference frame is attached to the object, then it should be on the center of gravity of the object, and if the reference frame is neither, the 0 potential should be at the center of gravity of the object+earth system.




Also another question that was already discussed in SE (but i didn't
find my answer there) is why do we talk about the potential energy of
a system (ball+Earth) but kinetic energy of an object (ball)?




Because you need to define what type of potential energy we are talking about, in this case it's gravity (could be electrical for example)



We now see that potential energy is relative to a type of force (gravity), reference frame, and the final nail in the coffin of your definition of potential energy, is that it's not "stored in the object" . Potential energies depend on the spacial position of the object in the force field.



I hope all these elements will help you understand potential energy. Meanwhile your teacher probably gave you a simplified version of this that is enough to solve the problems you are supposed to solve for now. The main idea the teacher probably wants to convey is that energy is conserved, and that like voltage, we are more interested in differences in potential than absolute potentials.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    A moving object, due to its inertia, has an energy (that could do damage in case it impacts something :) We call that energy "Kinetic"

    A system (like the earth and ball) can potentially (if allowed) cause one or both objects to move because of the gravity (in this case).



    If the force of gravity is perpendicular to the surface where the ball meets the earth, that hard surface prevents the ball from going anywhere so we are saying that the ball can potentially gain no speed thus it has "no potential energy".



    This is where the confusion starts, because in fact that ball had potential energy (if only allowed to move). The only place where the ball could really have zero potential energy would be at the lowest point on the surface of the earh (or that of the surface of the ocean floor depending on the density of the ball)
    If we move the ball to the side making available a cliff where the ball could potentially roll down on, we realise that the ball (or the system) did have potential energy, we just said it didn't, for practical reasons.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Your answer seems to put a lot of importance on where we actually define potential energy to be $0$, when in fact the $0$ point is arbitrary.
      $endgroup$
      – Aaron Stevens
      Jan 23 at 16:28






    • 1




      $begingroup$
      But then I don't get what the potential energy represents or why do we even need it when we can choose the point where potential energy is 0, then the value of potential energy will be different for the same object (position) everytime we choose a different 0 point. Like for eg if you compared two velocities one with the magnitude 0m/s and one 10m/s you know the other object is faster, but with PE the same object in one position can have different magnitude/value depending on the zero point. So the value doen't represent anything in my opinion, or what does it represent?
      $endgroup$
      – Lauren Sin
      Jan 23 at 16:32






    • 1




      $begingroup$
      @LaurenSin What really matters is changes in potential energy. The absolute value of it does not matter. Also when comparing two objects you still use the same zero-point for each object, so it is still relevant to say which object has more potential energy that could be converted to kinetic energy should both objects end up with the same potential energy while they move. But the same thing will happen to an object going from potential energies of $5 rm J$ to $3 rm J$ as one going from $2 rm J$ to $0 rm J$
      $endgroup$
      – Aaron Stevens
      Jan 23 at 16:36












    • $begingroup$
      You need to avoid these kind of abstract concepts, because they will lead you nowhere. Unless we have a machine which can convert gravitational energy into some other kind of energy, there is no practical value in calculating gravitational potential energy. Actually, we do have such a machine, it's called a waterwheel. Hydroelectric dams use this principle too. You might find that in these more concrete examples, the frame of reference defines itself. The water has a defined volume and mass,and a defined distance to fall at 1G acceleration, to convert its potential energy into kinetic energy.
      $endgroup$
      – Ed999
      Jan 23 at 18:23



















    0












    $begingroup$

    When An object is on a cliff at rest then relative to ground ,it will have 0 kinetic energy.we can take the surface of the cliff at 0 potential. When it is on surface of cliff,Two forces are working on it, The gravitational force of earth and the normal force of surface.Net force is zero, So it is in equilibrium.As net force is zero,so change in it's kinetic energy or its potential energy is also zero ($F=-dU/dx$).



    When it is falling from the cliff ,we removed the normal force ,so net force is acting that is gravitational force. This will change its potential (kinetic)energy. We assumed the cliff at 0 potential energy ,so the force will decrease the potential energy in negative as wolframe johny said.



    The “energy” is just an abstract accounting tool that we use to describe the results of those interactions in terms of the work being done.



    We talk about the potential energy of system ,because we need force that will cause change in potential energy,but kinetic energy is just because of the relative motion of the object






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      If the ball was on top of a cliff, then it already had gravitational potential energy. This comes from the fact that if the ball fell off the cliff, if would gain kinetic energy (while losing the potential energy it had from being on top of the cliff).



      Gravitational potential energy is usually discussed with respect to a system because there needs to be other mass(es) exerting gravity on an object to give it said potential energy. However, there are other types of potential energy internal to an object (e.g., chemical potential energy), that has no reliance on an external system.



      The main thing to remember is that potential energy is how much kinetic energy an object could potentially exert. An object being on top of a cliff has potential energy since it has the potential to drop off the cliff and gain kinetic energy.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I think there's something missing here about the arbitrariness of the zero point of kinetic/potential energy. Whatever numerical value we choose to assign to the potential energy of a ball at the top of a cliff is rather meaningless. Suppose the ball falls to the bottom of the cliff, gaining a certain amount of KE before it lands. Now it turns out the ball landed on a trap door and falls through, gaining even more KE, even though in the frame of reference we started in, the ball didn't ever have that much PE. All that matters is the change in KE and PE, but not the actual numerical values.
        $endgroup$
        – Nuclear Wang
        Jan 23 at 20:39











      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "151"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: false,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: null,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f456150%2fhow-can-an-object-with-zero-potential-and-kinetic-energy-ever-move%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      6 Answers
      6






      active

      oldest

      votes








      6 Answers
      6






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6












      $begingroup$

      Where we define the potential energy to be $0$ in classical mechanics is arbitrary. All that matters is the change in potential energy. Since you are just learning this stuff I will assume you are in an algebra based physics class, so I will avoid using calculus here.



      Potential energies are nice because they tell us how much work is done by a conservative force. More specifically, the work done by a conservative force is given by
      $$W_{cons}=-Delta U$$
      where $U$ is the potential energy associated with that conservative force. This is useful because we also know that the net work done on an object determines its change in kinetic energy
      $$W_{net}=Delta K$$



      So, if we consider your case where we just have one conservative force acting on the object, we can conclude that
      $$Delta K=-Delta U$$



      And so we see here that the only thing that determines how the motion of our object changes is just the change in potential energy. If we define the zero-point to be at the top of the cliff, then as the object falls its kinetic energy will grow and its potential energy will decrease and be negative when it hits the bottom of the cliff. If we define the zero-point to be at the bottom of the cliff then as the object falls its kinetic energy will grow and its potential energy will decrease to $0$ when it hits the bottom of the cliff. In either case the same thing happens because we have the same change in potential energy.




      Also another question that was already discussed in SE (but i didn't find my answer there) is why do we talk about the potential energy of a system (ball+Earth) but kinetic energy of an object (ball)?




      Typically in introductory physics classes we just consider the ball being in a uniform gravitational field and we don't even consider the Earth. However if you want to include the Earth in your system then when the ball falls the Earth will actually move slightly upwards to meet the ball. It will still be the case though that the work done by gravity on each object is related to the change in the potential energy of each object, which results in a change of kinetic energy.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        "All that matters is the change in potential energy". That is true when calculating kinetic energy gained, but two balls at equal heights in the air, one above high ground an one above a cliff still have the same potential energy. Does a ball on top of a mountain have potential energy? Yes because the potential energy it accumulated while being lifted up there, even thou it is standing still.
        $endgroup$
        – Alex Doe
        Jan 23 at 17:51












      • $begingroup$
        @AlexDoe You can say that the potential energy increased as you lifted the ball to the top of the mountain, but if you set $h=0$ at the top then you can say the ball's potential energy is $0$. That is why it is better to say "the ball's potential energy increased"
        $endgroup$
        – Aaron Stevens
        Jan 23 at 18:40










      • $begingroup$
        I see your point. I think the opinion differences are in terminology. The way I was looking at it is like if I go to the mall and I set a limit as to how much I want to spend there (arbitrary). That doesn't mean that that's all the money (potential energy) that I have. Thanks for the input thou. Seems like I am a minority/alone on this side of the isle thou :)
        $endgroup$
        – Alex Doe
        Jan 23 at 19:10










      • $begingroup$
        @AlexDoe In your analogy it doesn't matter where you set your spending limit. You could spend more or less money than your limit and have "positive or negative" cash relative to that limit, but either way the new shirt you buy will change the amount of cash you have in your account by the same amount no matter what you decide your spending limit is.
        $endgroup$
        – Aaron Stevens
        Jan 23 at 19:14








      • 1




        $begingroup$
        @AlexDoe: I agree with you; I think it would be better to say "All that matters is differences in potential energy" (rather than specifically change). The "zero" is arbitrary, but you need to fix it once and stick with it for a whole calculation -- you can't just keep resetting your arbitrary zero and expect everything to work out.
        $endgroup$
        – ruakh
        Jan 23 at 20:04


















      6












      $begingroup$

      Where we define the potential energy to be $0$ in classical mechanics is arbitrary. All that matters is the change in potential energy. Since you are just learning this stuff I will assume you are in an algebra based physics class, so I will avoid using calculus here.



      Potential energies are nice because they tell us how much work is done by a conservative force. More specifically, the work done by a conservative force is given by
      $$W_{cons}=-Delta U$$
      where $U$ is the potential energy associated with that conservative force. This is useful because we also know that the net work done on an object determines its change in kinetic energy
      $$W_{net}=Delta K$$



      So, if we consider your case where we just have one conservative force acting on the object, we can conclude that
      $$Delta K=-Delta U$$



      And so we see here that the only thing that determines how the motion of our object changes is just the change in potential energy. If we define the zero-point to be at the top of the cliff, then as the object falls its kinetic energy will grow and its potential energy will decrease and be negative when it hits the bottom of the cliff. If we define the zero-point to be at the bottom of the cliff then as the object falls its kinetic energy will grow and its potential energy will decrease to $0$ when it hits the bottom of the cliff. In either case the same thing happens because we have the same change in potential energy.




      Also another question that was already discussed in SE (but i didn't find my answer there) is why do we talk about the potential energy of a system (ball+Earth) but kinetic energy of an object (ball)?




      Typically in introductory physics classes we just consider the ball being in a uniform gravitational field and we don't even consider the Earth. However if you want to include the Earth in your system then when the ball falls the Earth will actually move slightly upwards to meet the ball. It will still be the case though that the work done by gravity on each object is related to the change in the potential energy of each object, which results in a change of kinetic energy.






      share|cite|improve this answer









      $endgroup$









      • 1




        $begingroup$
        "All that matters is the change in potential energy". That is true when calculating kinetic energy gained, but two balls at equal heights in the air, one above high ground an one above a cliff still have the same potential energy. Does a ball on top of a mountain have potential energy? Yes because the potential energy it accumulated while being lifted up there, even thou it is standing still.
        $endgroup$
        – Alex Doe
        Jan 23 at 17:51












      • $begingroup$
        @AlexDoe You can say that the potential energy increased as you lifted the ball to the top of the mountain, but if you set $h=0$ at the top then you can say the ball's potential energy is $0$. That is why it is better to say "the ball's potential energy increased"
        $endgroup$
        – Aaron Stevens
        Jan 23 at 18:40










      • $begingroup$
        I see your point. I think the opinion differences are in terminology. The way I was looking at it is like if I go to the mall and I set a limit as to how much I want to spend there (arbitrary). That doesn't mean that that's all the money (potential energy) that I have. Thanks for the input thou. Seems like I am a minority/alone on this side of the isle thou :)
        $endgroup$
        – Alex Doe
        Jan 23 at 19:10










      • $begingroup$
        @AlexDoe In your analogy it doesn't matter where you set your spending limit. You could spend more or less money than your limit and have "positive or negative" cash relative to that limit, but either way the new shirt you buy will change the amount of cash you have in your account by the same amount no matter what you decide your spending limit is.
        $endgroup$
        – Aaron Stevens
        Jan 23 at 19:14








      • 1




        $begingroup$
        @AlexDoe: I agree with you; I think it would be better to say "All that matters is differences in potential energy" (rather than specifically change). The "zero" is arbitrary, but you need to fix it once and stick with it for a whole calculation -- you can't just keep resetting your arbitrary zero and expect everything to work out.
        $endgroup$
        – ruakh
        Jan 23 at 20:04
















      6












      6








      6





      $begingroup$

      Where we define the potential energy to be $0$ in classical mechanics is arbitrary. All that matters is the change in potential energy. Since you are just learning this stuff I will assume you are in an algebra based physics class, so I will avoid using calculus here.



      Potential energies are nice because they tell us how much work is done by a conservative force. More specifically, the work done by a conservative force is given by
      $$W_{cons}=-Delta U$$
      where $U$ is the potential energy associated with that conservative force. This is useful because we also know that the net work done on an object determines its change in kinetic energy
      $$W_{net}=Delta K$$



      So, if we consider your case where we just have one conservative force acting on the object, we can conclude that
      $$Delta K=-Delta U$$



      And so we see here that the only thing that determines how the motion of our object changes is just the change in potential energy. If we define the zero-point to be at the top of the cliff, then as the object falls its kinetic energy will grow and its potential energy will decrease and be negative when it hits the bottom of the cliff. If we define the zero-point to be at the bottom of the cliff then as the object falls its kinetic energy will grow and its potential energy will decrease to $0$ when it hits the bottom of the cliff. In either case the same thing happens because we have the same change in potential energy.




      Also another question that was already discussed in SE (but i didn't find my answer there) is why do we talk about the potential energy of a system (ball+Earth) but kinetic energy of an object (ball)?




      Typically in introductory physics classes we just consider the ball being in a uniform gravitational field and we don't even consider the Earth. However if you want to include the Earth in your system then when the ball falls the Earth will actually move slightly upwards to meet the ball. It will still be the case though that the work done by gravity on each object is related to the change in the potential energy of each object, which results in a change of kinetic energy.






      share|cite|improve this answer









      $endgroup$



      Where we define the potential energy to be $0$ in classical mechanics is arbitrary. All that matters is the change in potential energy. Since you are just learning this stuff I will assume you are in an algebra based physics class, so I will avoid using calculus here.



      Potential energies are nice because they tell us how much work is done by a conservative force. More specifically, the work done by a conservative force is given by
      $$W_{cons}=-Delta U$$
      where $U$ is the potential energy associated with that conservative force. This is useful because we also know that the net work done on an object determines its change in kinetic energy
      $$W_{net}=Delta K$$



      So, if we consider your case where we just have one conservative force acting on the object, we can conclude that
      $$Delta K=-Delta U$$



      And so we see here that the only thing that determines how the motion of our object changes is just the change in potential energy. If we define the zero-point to be at the top of the cliff, then as the object falls its kinetic energy will grow and its potential energy will decrease and be negative when it hits the bottom of the cliff. If we define the zero-point to be at the bottom of the cliff then as the object falls its kinetic energy will grow and its potential energy will decrease to $0$ when it hits the bottom of the cliff. In either case the same thing happens because we have the same change in potential energy.




      Also another question that was already discussed in SE (but i didn't find my answer there) is why do we talk about the potential energy of a system (ball+Earth) but kinetic energy of an object (ball)?




      Typically in introductory physics classes we just consider the ball being in a uniform gravitational field and we don't even consider the Earth. However if you want to include the Earth in your system then when the ball falls the Earth will actually move slightly upwards to meet the ball. It will still be the case though that the work done by gravity on each object is related to the change in the potential energy of each object, which results in a change of kinetic energy.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 23 at 16:45









      Aaron StevensAaron Stevens

      10.2k31741




      10.2k31741








      • 1




        $begingroup$
        "All that matters is the change in potential energy". That is true when calculating kinetic energy gained, but two balls at equal heights in the air, one above high ground an one above a cliff still have the same potential energy. Does a ball on top of a mountain have potential energy? Yes because the potential energy it accumulated while being lifted up there, even thou it is standing still.
        $endgroup$
        – Alex Doe
        Jan 23 at 17:51












      • $begingroup$
        @AlexDoe You can say that the potential energy increased as you lifted the ball to the top of the mountain, but if you set $h=0$ at the top then you can say the ball's potential energy is $0$. That is why it is better to say "the ball's potential energy increased"
        $endgroup$
        – Aaron Stevens
        Jan 23 at 18:40










      • $begingroup$
        I see your point. I think the opinion differences are in terminology. The way I was looking at it is like if I go to the mall and I set a limit as to how much I want to spend there (arbitrary). That doesn't mean that that's all the money (potential energy) that I have. Thanks for the input thou. Seems like I am a minority/alone on this side of the isle thou :)
        $endgroup$
        – Alex Doe
        Jan 23 at 19:10










      • $begingroup$
        @AlexDoe In your analogy it doesn't matter where you set your spending limit. You could spend more or less money than your limit and have "positive or negative" cash relative to that limit, but either way the new shirt you buy will change the amount of cash you have in your account by the same amount no matter what you decide your spending limit is.
        $endgroup$
        – Aaron Stevens
        Jan 23 at 19:14








      • 1




        $begingroup$
        @AlexDoe: I agree with you; I think it would be better to say "All that matters is differences in potential energy" (rather than specifically change). The "zero" is arbitrary, but you need to fix it once and stick with it for a whole calculation -- you can't just keep resetting your arbitrary zero and expect everything to work out.
        $endgroup$
        – ruakh
        Jan 23 at 20:04
















      • 1




        $begingroup$
        "All that matters is the change in potential energy". That is true when calculating kinetic energy gained, but two balls at equal heights in the air, one above high ground an one above a cliff still have the same potential energy. Does a ball on top of a mountain have potential energy? Yes because the potential energy it accumulated while being lifted up there, even thou it is standing still.
        $endgroup$
        – Alex Doe
        Jan 23 at 17:51












      • $begingroup$
        @AlexDoe You can say that the potential energy increased as you lifted the ball to the top of the mountain, but if you set $h=0$ at the top then you can say the ball's potential energy is $0$. That is why it is better to say "the ball's potential energy increased"
        $endgroup$
        – Aaron Stevens
        Jan 23 at 18:40










      • $begingroup$
        I see your point. I think the opinion differences are in terminology. The way I was looking at it is like if I go to the mall and I set a limit as to how much I want to spend there (arbitrary). That doesn't mean that that's all the money (potential energy) that I have. Thanks for the input thou. Seems like I am a minority/alone on this side of the isle thou :)
        $endgroup$
        – Alex Doe
        Jan 23 at 19:10










      • $begingroup$
        @AlexDoe In your analogy it doesn't matter where you set your spending limit. You could spend more or less money than your limit and have "positive or negative" cash relative to that limit, but either way the new shirt you buy will change the amount of cash you have in your account by the same amount no matter what you decide your spending limit is.
        $endgroup$
        – Aaron Stevens
        Jan 23 at 19:14








      • 1




        $begingroup$
        @AlexDoe: I agree with you; I think it would be better to say "All that matters is differences in potential energy" (rather than specifically change). The "zero" is arbitrary, but you need to fix it once and stick with it for a whole calculation -- you can't just keep resetting your arbitrary zero and expect everything to work out.
        $endgroup$
        – ruakh
        Jan 23 at 20:04










      1




      1




      $begingroup$
      "All that matters is the change in potential energy". That is true when calculating kinetic energy gained, but two balls at equal heights in the air, one above high ground an one above a cliff still have the same potential energy. Does a ball on top of a mountain have potential energy? Yes because the potential energy it accumulated while being lifted up there, even thou it is standing still.
      $endgroup$
      – Alex Doe
      Jan 23 at 17:51






      $begingroup$
      "All that matters is the change in potential energy". That is true when calculating kinetic energy gained, but two balls at equal heights in the air, one above high ground an one above a cliff still have the same potential energy. Does a ball on top of a mountain have potential energy? Yes because the potential energy it accumulated while being lifted up there, even thou it is standing still.
      $endgroup$
      – Alex Doe
      Jan 23 at 17:51














      $begingroup$
      @AlexDoe You can say that the potential energy increased as you lifted the ball to the top of the mountain, but if you set $h=0$ at the top then you can say the ball's potential energy is $0$. That is why it is better to say "the ball's potential energy increased"
      $endgroup$
      – Aaron Stevens
      Jan 23 at 18:40




      $begingroup$
      @AlexDoe You can say that the potential energy increased as you lifted the ball to the top of the mountain, but if you set $h=0$ at the top then you can say the ball's potential energy is $0$. That is why it is better to say "the ball's potential energy increased"
      $endgroup$
      – Aaron Stevens
      Jan 23 at 18:40












      $begingroup$
      I see your point. I think the opinion differences are in terminology. The way I was looking at it is like if I go to the mall and I set a limit as to how much I want to spend there (arbitrary). That doesn't mean that that's all the money (potential energy) that I have. Thanks for the input thou. Seems like I am a minority/alone on this side of the isle thou :)
      $endgroup$
      – Alex Doe
      Jan 23 at 19:10




      $begingroup$
      I see your point. I think the opinion differences are in terminology. The way I was looking at it is like if I go to the mall and I set a limit as to how much I want to spend there (arbitrary). That doesn't mean that that's all the money (potential energy) that I have. Thanks for the input thou. Seems like I am a minority/alone on this side of the isle thou :)
      $endgroup$
      – Alex Doe
      Jan 23 at 19:10












      $begingroup$
      @AlexDoe In your analogy it doesn't matter where you set your spending limit. You could spend more or less money than your limit and have "positive or negative" cash relative to that limit, but either way the new shirt you buy will change the amount of cash you have in your account by the same amount no matter what you decide your spending limit is.
      $endgroup$
      – Aaron Stevens
      Jan 23 at 19:14






      $begingroup$
      @AlexDoe In your analogy it doesn't matter where you set your spending limit. You could spend more or less money than your limit and have "positive or negative" cash relative to that limit, but either way the new shirt you buy will change the amount of cash you have in your account by the same amount no matter what you decide your spending limit is.
      $endgroup$
      – Aaron Stevens
      Jan 23 at 19:14






      1




      1




      $begingroup$
      @AlexDoe: I agree with you; I think it would be better to say "All that matters is differences in potential energy" (rather than specifically change). The "zero" is arbitrary, but you need to fix it once and stick with it for a whole calculation -- you can't just keep resetting your arbitrary zero and expect everything to work out.
      $endgroup$
      – ruakh
      Jan 23 at 20:04






      $begingroup$
      @AlexDoe: I agree with you; I think it would be better to say "All that matters is differences in potential energy" (rather than specifically change). The "zero" is arbitrary, but you need to fix it once and stick with it for a whole calculation -- you can't just keep resetting your arbitrary zero and expect everything to work out.
      $endgroup$
      – ruakh
      Jan 23 at 20:04













      4












      $begingroup$

      The energy of an object consists of its internal energy (kinetic and potential energy possessed internally to an object at the atomic or molecular level) and external kinetic and potential energy. Its external kinetic and potential energy is always with respect to some external frame of reference.



      The ball on the surface of the earth may have zero potential energy with respect to the surface of the earth, but has gravitational potential energy with respect to the center of the earth. It doesn’t move because the force of gravity downward on the ball is equal and opposite to the force exerted upward on it by the ground. But imagine if there was a trench that went from one side of the earth to the other. If the ball were positioned over the hole and allowed to fall, it would accelerate toward the center of the earth losing potential energy and gaining kinetic energy, the latter being a maximum at the center where its potential energy is zero. Then it would decelerate as it goes past the center as it gains potential energy and loses kinetic energy, ultimately stopping at the other side (ignoring air resistance).



      Even the kinetic energy of the object is relative to the frame of reference where the velocity is measured. If you are in a car moving at velocity $v$ with respect to the road, it has zero kinetic energy in your frame of reference. It is not “moving” in your frame of reference, but is moving in the frame of reference of a person standing on the road. To that person the car has kinetic energy. The person standing on the road is “moving” with respect to your frame of reference in the car, so that person has kinetic energy with respect to you in the car. The person standing on the road has kinetic energy with respect to a frame of reference other than the earth (since the earth is rotating). And so on…



      Bottom line, everything can be considered to have kinetic energy (be moving) with respect to some frame of reference and have potential energy (due to its position) with respect to some frame of reference.



      Hope this help.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I like how you mention the arbitrariness of the "zero-point" of kinetic energy. I knew this was the case, but never realized how they are kind of the same idea. In a sense, just like how we define where $h=0$ for the zero-point of potential energy, we can also define what constitutes as $v=0$ for a sort of "zero-point" of kinetic energy.
        $endgroup$
        – Aaron Stevens
        Jan 23 at 16:56










      • $begingroup$
        Thanks. I like the way you call it "zero point" kinetic and potential energy.
        $endgroup$
        – Bob D
        Jan 23 at 17:02










      • $begingroup$
        Yeah it is probably not the most accurate terminology considering the use of "zero-point energy" in QM, but since we are in the classical realm for this question I think I will keep it haha
        $endgroup$
        – Aaron Stevens
        Jan 23 at 17:03






      • 2




        $begingroup$
        @Ed999 Bob never said anything contradicting what you just explained
        $endgroup$
        – Aaron Stevens
        Jan 23 at 18:44






      • 1




        $begingroup$
        @Ed999 I thought that's what I said.
        $endgroup$
        – Bob D
        Jan 23 at 19:18
















      4












      $begingroup$

      The energy of an object consists of its internal energy (kinetic and potential energy possessed internally to an object at the atomic or molecular level) and external kinetic and potential energy. Its external kinetic and potential energy is always with respect to some external frame of reference.



      The ball on the surface of the earth may have zero potential energy with respect to the surface of the earth, but has gravitational potential energy with respect to the center of the earth. It doesn’t move because the force of gravity downward on the ball is equal and opposite to the force exerted upward on it by the ground. But imagine if there was a trench that went from one side of the earth to the other. If the ball were positioned over the hole and allowed to fall, it would accelerate toward the center of the earth losing potential energy and gaining kinetic energy, the latter being a maximum at the center where its potential energy is zero. Then it would decelerate as it goes past the center as it gains potential energy and loses kinetic energy, ultimately stopping at the other side (ignoring air resistance).



      Even the kinetic energy of the object is relative to the frame of reference where the velocity is measured. If you are in a car moving at velocity $v$ with respect to the road, it has zero kinetic energy in your frame of reference. It is not “moving” in your frame of reference, but is moving in the frame of reference of a person standing on the road. To that person the car has kinetic energy. The person standing on the road is “moving” with respect to your frame of reference in the car, so that person has kinetic energy with respect to you in the car. The person standing on the road has kinetic energy with respect to a frame of reference other than the earth (since the earth is rotating). And so on…



      Bottom line, everything can be considered to have kinetic energy (be moving) with respect to some frame of reference and have potential energy (due to its position) with respect to some frame of reference.



      Hope this help.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        I like how you mention the arbitrariness of the "zero-point" of kinetic energy. I knew this was the case, but never realized how they are kind of the same idea. In a sense, just like how we define where $h=0$ for the zero-point of potential energy, we can also define what constitutes as $v=0$ for a sort of "zero-point" of kinetic energy.
        $endgroup$
        – Aaron Stevens
        Jan 23 at 16:56










      • $begingroup$
        Thanks. I like the way you call it "zero point" kinetic and potential energy.
        $endgroup$
        – Bob D
        Jan 23 at 17:02










      • $begingroup$
        Yeah it is probably not the most accurate terminology considering the use of "zero-point energy" in QM, but since we are in the classical realm for this question I think I will keep it haha
        $endgroup$
        – Aaron Stevens
        Jan 23 at 17:03






      • 2




        $begingroup$
        @Ed999 Bob never said anything contradicting what you just explained
        $endgroup$
        – Aaron Stevens
        Jan 23 at 18:44






      • 1




        $begingroup$
        @Ed999 I thought that's what I said.
        $endgroup$
        – Bob D
        Jan 23 at 19:18














      4












      4








      4





      $begingroup$

      The energy of an object consists of its internal energy (kinetic and potential energy possessed internally to an object at the atomic or molecular level) and external kinetic and potential energy. Its external kinetic and potential energy is always with respect to some external frame of reference.



      The ball on the surface of the earth may have zero potential energy with respect to the surface of the earth, but has gravitational potential energy with respect to the center of the earth. It doesn’t move because the force of gravity downward on the ball is equal and opposite to the force exerted upward on it by the ground. But imagine if there was a trench that went from one side of the earth to the other. If the ball were positioned over the hole and allowed to fall, it would accelerate toward the center of the earth losing potential energy and gaining kinetic energy, the latter being a maximum at the center where its potential energy is zero. Then it would decelerate as it goes past the center as it gains potential energy and loses kinetic energy, ultimately stopping at the other side (ignoring air resistance).



      Even the kinetic energy of the object is relative to the frame of reference where the velocity is measured. If you are in a car moving at velocity $v$ with respect to the road, it has zero kinetic energy in your frame of reference. It is not “moving” in your frame of reference, but is moving in the frame of reference of a person standing on the road. To that person the car has kinetic energy. The person standing on the road is “moving” with respect to your frame of reference in the car, so that person has kinetic energy with respect to you in the car. The person standing on the road has kinetic energy with respect to a frame of reference other than the earth (since the earth is rotating). And so on…



      Bottom line, everything can be considered to have kinetic energy (be moving) with respect to some frame of reference and have potential energy (due to its position) with respect to some frame of reference.



      Hope this help.






      share|cite|improve this answer









      $endgroup$



      The energy of an object consists of its internal energy (kinetic and potential energy possessed internally to an object at the atomic or molecular level) and external kinetic and potential energy. Its external kinetic and potential energy is always with respect to some external frame of reference.



      The ball on the surface of the earth may have zero potential energy with respect to the surface of the earth, but has gravitational potential energy with respect to the center of the earth. It doesn’t move because the force of gravity downward on the ball is equal and opposite to the force exerted upward on it by the ground. But imagine if there was a trench that went from one side of the earth to the other. If the ball were positioned over the hole and allowed to fall, it would accelerate toward the center of the earth losing potential energy and gaining kinetic energy, the latter being a maximum at the center where its potential energy is zero. Then it would decelerate as it goes past the center as it gains potential energy and loses kinetic energy, ultimately stopping at the other side (ignoring air resistance).



      Even the kinetic energy of the object is relative to the frame of reference where the velocity is measured. If you are in a car moving at velocity $v$ with respect to the road, it has zero kinetic energy in your frame of reference. It is not “moving” in your frame of reference, but is moving in the frame of reference of a person standing on the road. To that person the car has kinetic energy. The person standing on the road is “moving” with respect to your frame of reference in the car, so that person has kinetic energy with respect to you in the car. The person standing on the road has kinetic energy with respect to a frame of reference other than the earth (since the earth is rotating). And so on…



      Bottom line, everything can be considered to have kinetic energy (be moving) with respect to some frame of reference and have potential energy (due to its position) with respect to some frame of reference.



      Hope this help.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 23 at 16:48









      Bob DBob D

      2,7671214




      2,7671214












      • $begingroup$
        I like how you mention the arbitrariness of the "zero-point" of kinetic energy. I knew this was the case, but never realized how they are kind of the same idea. In a sense, just like how we define where $h=0$ for the zero-point of potential energy, we can also define what constitutes as $v=0$ for a sort of "zero-point" of kinetic energy.
        $endgroup$
        – Aaron Stevens
        Jan 23 at 16:56










      • $begingroup$
        Thanks. I like the way you call it "zero point" kinetic and potential energy.
        $endgroup$
        – Bob D
        Jan 23 at 17:02










      • $begingroup$
        Yeah it is probably not the most accurate terminology considering the use of "zero-point energy" in QM, but since we are in the classical realm for this question I think I will keep it haha
        $endgroup$
        – Aaron Stevens
        Jan 23 at 17:03






      • 2




        $begingroup$
        @Ed999 Bob never said anything contradicting what you just explained
        $endgroup$
        – Aaron Stevens
        Jan 23 at 18:44






      • 1




        $begingroup$
        @Ed999 I thought that's what I said.
        $endgroup$
        – Bob D
        Jan 23 at 19:18


















      • $begingroup$
        I like how you mention the arbitrariness of the "zero-point" of kinetic energy. I knew this was the case, but never realized how they are kind of the same idea. In a sense, just like how we define where $h=0$ for the zero-point of potential energy, we can also define what constitutes as $v=0$ for a sort of "zero-point" of kinetic energy.
        $endgroup$
        – Aaron Stevens
        Jan 23 at 16:56










      • $begingroup$
        Thanks. I like the way you call it "zero point" kinetic and potential energy.
        $endgroup$
        – Bob D
        Jan 23 at 17:02










      • $begingroup$
        Yeah it is probably not the most accurate terminology considering the use of "zero-point energy" in QM, but since we are in the classical realm for this question I think I will keep it haha
        $endgroup$
        – Aaron Stevens
        Jan 23 at 17:03






      • 2




        $begingroup$
        @Ed999 Bob never said anything contradicting what you just explained
        $endgroup$
        – Aaron Stevens
        Jan 23 at 18:44






      • 1




        $begingroup$
        @Ed999 I thought that's what I said.
        $endgroup$
        – Bob D
        Jan 23 at 19:18
















      $begingroup$
      I like how you mention the arbitrariness of the "zero-point" of kinetic energy. I knew this was the case, but never realized how they are kind of the same idea. In a sense, just like how we define where $h=0$ for the zero-point of potential energy, we can also define what constitutes as $v=0$ for a sort of "zero-point" of kinetic energy.
      $endgroup$
      – Aaron Stevens
      Jan 23 at 16:56




      $begingroup$
      I like how you mention the arbitrariness of the "zero-point" of kinetic energy. I knew this was the case, but never realized how they are kind of the same idea. In a sense, just like how we define where $h=0$ for the zero-point of potential energy, we can also define what constitutes as $v=0$ for a sort of "zero-point" of kinetic energy.
      $endgroup$
      – Aaron Stevens
      Jan 23 at 16:56












      $begingroup$
      Thanks. I like the way you call it "zero point" kinetic and potential energy.
      $endgroup$
      – Bob D
      Jan 23 at 17:02




      $begingroup$
      Thanks. I like the way you call it "zero point" kinetic and potential energy.
      $endgroup$
      – Bob D
      Jan 23 at 17:02












      $begingroup$
      Yeah it is probably not the most accurate terminology considering the use of "zero-point energy" in QM, but since we are in the classical realm for this question I think I will keep it haha
      $endgroup$
      – Aaron Stevens
      Jan 23 at 17:03




      $begingroup$
      Yeah it is probably not the most accurate terminology considering the use of "zero-point energy" in QM, but since we are in the classical realm for this question I think I will keep it haha
      $endgroup$
      – Aaron Stevens
      Jan 23 at 17:03




      2




      2




      $begingroup$
      @Ed999 Bob never said anything contradicting what you just explained
      $endgroup$
      – Aaron Stevens
      Jan 23 at 18:44




      $begingroup$
      @Ed999 Bob never said anything contradicting what you just explained
      $endgroup$
      – Aaron Stevens
      Jan 23 at 18:44




      1




      1




      $begingroup$
      @Ed999 I thought that's what I said.
      $endgroup$
      – Bob D
      Jan 23 at 19:18




      $begingroup$
      @Ed999 I thought that's what I said.
      $endgroup$
      – Bob D
      Jan 23 at 19:18











      2












      $begingroup$


      I am not sure how to ask this question but I am learning about
      potential energy (high school physics) and from the deffinition of a
      potential energy (energy stored in an object with the potential to
      convert into other type of energy)




      this definition of potential energy, sounds more like the energy itself. More below




      i don't understand how for eg. when we have an object (let's say a
      ball) on the ground, it has zero kinetic energy and also zero
      potential energy and now let's say the ball starts falling of a cliff
      so it will be gaining kinetic energy but when it had no potential
      energy, how can it be now gaining energy?




      indeed, in the planet+object system, the true 0 potential, would be better put at the center of gravity the earth if the reference frame is attached to the earth, if the reference frame is attached to the object, then it should be on the center of gravity of the object, and if the reference frame is neither, the 0 potential should be at the center of gravity of the object+earth system.




      Also another question that was already discussed in SE (but i didn't
      find my answer there) is why do we talk about the potential energy of
      a system (ball+Earth) but kinetic energy of an object (ball)?




      Because you need to define what type of potential energy we are talking about, in this case it's gravity (could be electrical for example)



      We now see that potential energy is relative to a type of force (gravity), reference frame, and the final nail in the coffin of your definition of potential energy, is that it's not "stored in the object" . Potential energies depend on the spacial position of the object in the force field.



      I hope all these elements will help you understand potential energy. Meanwhile your teacher probably gave you a simplified version of this that is enough to solve the problems you are supposed to solve for now. The main idea the teacher probably wants to convey is that energy is conserved, and that like voltage, we are more interested in differences in potential than absolute potentials.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$


        I am not sure how to ask this question but I am learning about
        potential energy (high school physics) and from the deffinition of a
        potential energy (energy stored in an object with the potential to
        convert into other type of energy)




        this definition of potential energy, sounds more like the energy itself. More below




        i don't understand how for eg. when we have an object (let's say a
        ball) on the ground, it has zero kinetic energy and also zero
        potential energy and now let's say the ball starts falling of a cliff
        so it will be gaining kinetic energy but when it had no potential
        energy, how can it be now gaining energy?




        indeed, in the planet+object system, the true 0 potential, would be better put at the center of gravity the earth if the reference frame is attached to the earth, if the reference frame is attached to the object, then it should be on the center of gravity of the object, and if the reference frame is neither, the 0 potential should be at the center of gravity of the object+earth system.




        Also another question that was already discussed in SE (but i didn't
        find my answer there) is why do we talk about the potential energy of
        a system (ball+Earth) but kinetic energy of an object (ball)?




        Because you need to define what type of potential energy we are talking about, in this case it's gravity (could be electrical for example)



        We now see that potential energy is relative to a type of force (gravity), reference frame, and the final nail in the coffin of your definition of potential energy, is that it's not "stored in the object" . Potential energies depend on the spacial position of the object in the force field.



        I hope all these elements will help you understand potential energy. Meanwhile your teacher probably gave you a simplified version of this that is enough to solve the problems you are supposed to solve for now. The main idea the teacher probably wants to convey is that energy is conserved, and that like voltage, we are more interested in differences in potential than absolute potentials.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$


          I am not sure how to ask this question but I am learning about
          potential energy (high school physics) and from the deffinition of a
          potential energy (energy stored in an object with the potential to
          convert into other type of energy)




          this definition of potential energy, sounds more like the energy itself. More below




          i don't understand how for eg. when we have an object (let's say a
          ball) on the ground, it has zero kinetic energy and also zero
          potential energy and now let's say the ball starts falling of a cliff
          so it will be gaining kinetic energy but when it had no potential
          energy, how can it be now gaining energy?




          indeed, in the planet+object system, the true 0 potential, would be better put at the center of gravity the earth if the reference frame is attached to the earth, if the reference frame is attached to the object, then it should be on the center of gravity of the object, and if the reference frame is neither, the 0 potential should be at the center of gravity of the object+earth system.




          Also another question that was already discussed in SE (but i didn't
          find my answer there) is why do we talk about the potential energy of
          a system (ball+Earth) but kinetic energy of an object (ball)?




          Because you need to define what type of potential energy we are talking about, in this case it's gravity (could be electrical for example)



          We now see that potential energy is relative to a type of force (gravity), reference frame, and the final nail in the coffin of your definition of potential energy, is that it's not "stored in the object" . Potential energies depend on the spacial position of the object in the force field.



          I hope all these elements will help you understand potential energy. Meanwhile your teacher probably gave you a simplified version of this that is enough to solve the problems you are supposed to solve for now. The main idea the teacher probably wants to convey is that energy is conserved, and that like voltage, we are more interested in differences in potential than absolute potentials.






          share|cite|improve this answer









          $endgroup$




          I am not sure how to ask this question but I am learning about
          potential energy (high school physics) and from the deffinition of a
          potential energy (energy stored in an object with the potential to
          convert into other type of energy)




          this definition of potential energy, sounds more like the energy itself. More below




          i don't understand how for eg. when we have an object (let's say a
          ball) on the ground, it has zero kinetic energy and also zero
          potential energy and now let's say the ball starts falling of a cliff
          so it will be gaining kinetic energy but when it had no potential
          energy, how can it be now gaining energy?




          indeed, in the planet+object system, the true 0 potential, would be better put at the center of gravity the earth if the reference frame is attached to the earth, if the reference frame is attached to the object, then it should be on the center of gravity of the object, and if the reference frame is neither, the 0 potential should be at the center of gravity of the object+earth system.




          Also another question that was already discussed in SE (but i didn't
          find my answer there) is why do we talk about the potential energy of
          a system (ball+Earth) but kinetic energy of an object (ball)?




          Because you need to define what type of potential energy we are talking about, in this case it's gravity (could be electrical for example)



          We now see that potential energy is relative to a type of force (gravity), reference frame, and the final nail in the coffin of your definition of potential energy, is that it's not "stored in the object" . Potential energies depend on the spacial position of the object in the force field.



          I hope all these elements will help you understand potential energy. Meanwhile your teacher probably gave you a simplified version of this that is enough to solve the problems you are supposed to solve for now. The main idea the teacher probably wants to convey is that energy is conserved, and that like voltage, we are more interested in differences in potential than absolute potentials.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 17:22









          Manu de HanoiManu de Hanoi

          487212




          487212























              1












              $begingroup$

              A moving object, due to its inertia, has an energy (that could do damage in case it impacts something :) We call that energy "Kinetic"

              A system (like the earth and ball) can potentially (if allowed) cause one or both objects to move because of the gravity (in this case).



              If the force of gravity is perpendicular to the surface where the ball meets the earth, that hard surface prevents the ball from going anywhere so we are saying that the ball can potentially gain no speed thus it has "no potential energy".



              This is where the confusion starts, because in fact that ball had potential energy (if only allowed to move). The only place where the ball could really have zero potential energy would be at the lowest point on the surface of the earh (or that of the surface of the ocean floor depending on the density of the ball)
              If we move the ball to the side making available a cliff where the ball could potentially roll down on, we realise that the ball (or the system) did have potential energy, we just said it didn't, for practical reasons.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                Your answer seems to put a lot of importance on where we actually define potential energy to be $0$, when in fact the $0$ point is arbitrary.
                $endgroup$
                – Aaron Stevens
                Jan 23 at 16:28






              • 1




                $begingroup$
                But then I don't get what the potential energy represents or why do we even need it when we can choose the point where potential energy is 0, then the value of potential energy will be different for the same object (position) everytime we choose a different 0 point. Like for eg if you compared two velocities one with the magnitude 0m/s and one 10m/s you know the other object is faster, but with PE the same object in one position can have different magnitude/value depending on the zero point. So the value doen't represent anything in my opinion, or what does it represent?
                $endgroup$
                – Lauren Sin
                Jan 23 at 16:32






              • 1




                $begingroup$
                @LaurenSin What really matters is changes in potential energy. The absolute value of it does not matter. Also when comparing two objects you still use the same zero-point for each object, so it is still relevant to say which object has more potential energy that could be converted to kinetic energy should both objects end up with the same potential energy while they move. But the same thing will happen to an object going from potential energies of $5 rm J$ to $3 rm J$ as one going from $2 rm J$ to $0 rm J$
                $endgroup$
                – Aaron Stevens
                Jan 23 at 16:36












              • $begingroup$
                You need to avoid these kind of abstract concepts, because they will lead you nowhere. Unless we have a machine which can convert gravitational energy into some other kind of energy, there is no practical value in calculating gravitational potential energy. Actually, we do have such a machine, it's called a waterwheel. Hydroelectric dams use this principle too. You might find that in these more concrete examples, the frame of reference defines itself. The water has a defined volume and mass,and a defined distance to fall at 1G acceleration, to convert its potential energy into kinetic energy.
                $endgroup$
                – Ed999
                Jan 23 at 18:23
















              1












              $begingroup$

              A moving object, due to its inertia, has an energy (that could do damage in case it impacts something :) We call that energy "Kinetic"

              A system (like the earth and ball) can potentially (if allowed) cause one or both objects to move because of the gravity (in this case).



              If the force of gravity is perpendicular to the surface where the ball meets the earth, that hard surface prevents the ball from going anywhere so we are saying that the ball can potentially gain no speed thus it has "no potential energy".



              This is where the confusion starts, because in fact that ball had potential energy (if only allowed to move). The only place where the ball could really have zero potential energy would be at the lowest point on the surface of the earh (or that of the surface of the ocean floor depending on the density of the ball)
              If we move the ball to the side making available a cliff where the ball could potentially roll down on, we realise that the ball (or the system) did have potential energy, we just said it didn't, for practical reasons.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                Your answer seems to put a lot of importance on where we actually define potential energy to be $0$, when in fact the $0$ point is arbitrary.
                $endgroup$
                – Aaron Stevens
                Jan 23 at 16:28






              • 1




                $begingroup$
                But then I don't get what the potential energy represents or why do we even need it when we can choose the point where potential energy is 0, then the value of potential energy will be different for the same object (position) everytime we choose a different 0 point. Like for eg if you compared two velocities one with the magnitude 0m/s and one 10m/s you know the other object is faster, but with PE the same object in one position can have different magnitude/value depending on the zero point. So the value doen't represent anything in my opinion, or what does it represent?
                $endgroup$
                – Lauren Sin
                Jan 23 at 16:32






              • 1




                $begingroup$
                @LaurenSin What really matters is changes in potential energy. The absolute value of it does not matter. Also when comparing two objects you still use the same zero-point for each object, so it is still relevant to say which object has more potential energy that could be converted to kinetic energy should both objects end up with the same potential energy while they move. But the same thing will happen to an object going from potential energies of $5 rm J$ to $3 rm J$ as one going from $2 rm J$ to $0 rm J$
                $endgroup$
                – Aaron Stevens
                Jan 23 at 16:36












              • $begingroup$
                You need to avoid these kind of abstract concepts, because they will lead you nowhere. Unless we have a machine which can convert gravitational energy into some other kind of energy, there is no practical value in calculating gravitational potential energy. Actually, we do have such a machine, it's called a waterwheel. Hydroelectric dams use this principle too. You might find that in these more concrete examples, the frame of reference defines itself. The water has a defined volume and mass,and a defined distance to fall at 1G acceleration, to convert its potential energy into kinetic energy.
                $endgroup$
                – Ed999
                Jan 23 at 18:23














              1












              1








              1





              $begingroup$

              A moving object, due to its inertia, has an energy (that could do damage in case it impacts something :) We call that energy "Kinetic"

              A system (like the earth and ball) can potentially (if allowed) cause one or both objects to move because of the gravity (in this case).



              If the force of gravity is perpendicular to the surface where the ball meets the earth, that hard surface prevents the ball from going anywhere so we are saying that the ball can potentially gain no speed thus it has "no potential energy".



              This is where the confusion starts, because in fact that ball had potential energy (if only allowed to move). The only place where the ball could really have zero potential energy would be at the lowest point on the surface of the earh (or that of the surface of the ocean floor depending on the density of the ball)
              If we move the ball to the side making available a cliff where the ball could potentially roll down on, we realise that the ball (or the system) did have potential energy, we just said it didn't, for practical reasons.






              share|cite|improve this answer











              $endgroup$



              A moving object, due to its inertia, has an energy (that could do damage in case it impacts something :) We call that energy "Kinetic"

              A system (like the earth and ball) can potentially (if allowed) cause one or both objects to move because of the gravity (in this case).



              If the force of gravity is perpendicular to the surface where the ball meets the earth, that hard surface prevents the ball from going anywhere so we are saying that the ball can potentially gain no speed thus it has "no potential energy".



              This is where the confusion starts, because in fact that ball had potential energy (if only allowed to move). The only place where the ball could really have zero potential energy would be at the lowest point on the surface of the earh (or that of the surface of the ocean floor depending on the density of the ball)
              If we move the ball to the side making available a cliff where the ball could potentially roll down on, we realise that the ball (or the system) did have potential energy, we just said it didn't, for practical reasons.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 23 at 16:17

























              answered Jan 23 at 15:55









              Alex DoeAlex Doe

              823312




              823312








              • 1




                $begingroup$
                Your answer seems to put a lot of importance on where we actually define potential energy to be $0$, when in fact the $0$ point is arbitrary.
                $endgroup$
                – Aaron Stevens
                Jan 23 at 16:28






              • 1




                $begingroup$
                But then I don't get what the potential energy represents or why do we even need it when we can choose the point where potential energy is 0, then the value of potential energy will be different for the same object (position) everytime we choose a different 0 point. Like for eg if you compared two velocities one with the magnitude 0m/s and one 10m/s you know the other object is faster, but with PE the same object in one position can have different magnitude/value depending on the zero point. So the value doen't represent anything in my opinion, or what does it represent?
                $endgroup$
                – Lauren Sin
                Jan 23 at 16:32






              • 1




                $begingroup$
                @LaurenSin What really matters is changes in potential energy. The absolute value of it does not matter. Also when comparing two objects you still use the same zero-point for each object, so it is still relevant to say which object has more potential energy that could be converted to kinetic energy should both objects end up with the same potential energy while they move. But the same thing will happen to an object going from potential energies of $5 rm J$ to $3 rm J$ as one going from $2 rm J$ to $0 rm J$
                $endgroup$
                – Aaron Stevens
                Jan 23 at 16:36












              • $begingroup$
                You need to avoid these kind of abstract concepts, because they will lead you nowhere. Unless we have a machine which can convert gravitational energy into some other kind of energy, there is no practical value in calculating gravitational potential energy. Actually, we do have such a machine, it's called a waterwheel. Hydroelectric dams use this principle too. You might find that in these more concrete examples, the frame of reference defines itself. The water has a defined volume and mass,and a defined distance to fall at 1G acceleration, to convert its potential energy into kinetic energy.
                $endgroup$
                – Ed999
                Jan 23 at 18:23














              • 1




                $begingroup$
                Your answer seems to put a lot of importance on where we actually define potential energy to be $0$, when in fact the $0$ point is arbitrary.
                $endgroup$
                – Aaron Stevens
                Jan 23 at 16:28






              • 1




                $begingroup$
                But then I don't get what the potential energy represents or why do we even need it when we can choose the point where potential energy is 0, then the value of potential energy will be different for the same object (position) everytime we choose a different 0 point. Like for eg if you compared two velocities one with the magnitude 0m/s and one 10m/s you know the other object is faster, but with PE the same object in one position can have different magnitude/value depending on the zero point. So the value doen't represent anything in my opinion, or what does it represent?
                $endgroup$
                – Lauren Sin
                Jan 23 at 16:32






              • 1




                $begingroup$
                @LaurenSin What really matters is changes in potential energy. The absolute value of it does not matter. Also when comparing two objects you still use the same zero-point for each object, so it is still relevant to say which object has more potential energy that could be converted to kinetic energy should both objects end up with the same potential energy while they move. But the same thing will happen to an object going from potential energies of $5 rm J$ to $3 rm J$ as one going from $2 rm J$ to $0 rm J$
                $endgroup$
                – Aaron Stevens
                Jan 23 at 16:36












              • $begingroup$
                You need to avoid these kind of abstract concepts, because they will lead you nowhere. Unless we have a machine which can convert gravitational energy into some other kind of energy, there is no practical value in calculating gravitational potential energy. Actually, we do have such a machine, it's called a waterwheel. Hydroelectric dams use this principle too. You might find that in these more concrete examples, the frame of reference defines itself. The water has a defined volume and mass,and a defined distance to fall at 1G acceleration, to convert its potential energy into kinetic energy.
                $endgroup$
                – Ed999
                Jan 23 at 18:23








              1




              1




              $begingroup$
              Your answer seems to put a lot of importance on where we actually define potential energy to be $0$, when in fact the $0$ point is arbitrary.
              $endgroup$
              – Aaron Stevens
              Jan 23 at 16:28




              $begingroup$
              Your answer seems to put a lot of importance on where we actually define potential energy to be $0$, when in fact the $0$ point is arbitrary.
              $endgroup$
              – Aaron Stevens
              Jan 23 at 16:28




              1




              1




              $begingroup$
              But then I don't get what the potential energy represents or why do we even need it when we can choose the point where potential energy is 0, then the value of potential energy will be different for the same object (position) everytime we choose a different 0 point. Like for eg if you compared two velocities one with the magnitude 0m/s and one 10m/s you know the other object is faster, but with PE the same object in one position can have different magnitude/value depending on the zero point. So the value doen't represent anything in my opinion, or what does it represent?
              $endgroup$
              – Lauren Sin
              Jan 23 at 16:32




              $begingroup$
              But then I don't get what the potential energy represents or why do we even need it when we can choose the point where potential energy is 0, then the value of potential energy will be different for the same object (position) everytime we choose a different 0 point. Like for eg if you compared two velocities one with the magnitude 0m/s and one 10m/s you know the other object is faster, but with PE the same object in one position can have different magnitude/value depending on the zero point. So the value doen't represent anything in my opinion, or what does it represent?
              $endgroup$
              – Lauren Sin
              Jan 23 at 16:32




              1




              1




              $begingroup$
              @LaurenSin What really matters is changes in potential energy. The absolute value of it does not matter. Also when comparing two objects you still use the same zero-point for each object, so it is still relevant to say which object has more potential energy that could be converted to kinetic energy should both objects end up with the same potential energy while they move. But the same thing will happen to an object going from potential energies of $5 rm J$ to $3 rm J$ as one going from $2 rm J$ to $0 rm J$
              $endgroup$
              – Aaron Stevens
              Jan 23 at 16:36






              $begingroup$
              @LaurenSin What really matters is changes in potential energy. The absolute value of it does not matter. Also when comparing two objects you still use the same zero-point for each object, so it is still relevant to say which object has more potential energy that could be converted to kinetic energy should both objects end up with the same potential energy while they move. But the same thing will happen to an object going from potential energies of $5 rm J$ to $3 rm J$ as one going from $2 rm J$ to $0 rm J$
              $endgroup$
              – Aaron Stevens
              Jan 23 at 16:36














              $begingroup$
              You need to avoid these kind of abstract concepts, because they will lead you nowhere. Unless we have a machine which can convert gravitational energy into some other kind of energy, there is no practical value in calculating gravitational potential energy. Actually, we do have such a machine, it's called a waterwheel. Hydroelectric dams use this principle too. You might find that in these more concrete examples, the frame of reference defines itself. The water has a defined volume and mass,and a defined distance to fall at 1G acceleration, to convert its potential energy into kinetic energy.
              $endgroup$
              – Ed999
              Jan 23 at 18:23




              $begingroup$
              You need to avoid these kind of abstract concepts, because they will lead you nowhere. Unless we have a machine which can convert gravitational energy into some other kind of energy, there is no practical value in calculating gravitational potential energy. Actually, we do have such a machine, it's called a waterwheel. Hydroelectric dams use this principle too. You might find that in these more concrete examples, the frame of reference defines itself. The water has a defined volume and mass,and a defined distance to fall at 1G acceleration, to convert its potential energy into kinetic energy.
              $endgroup$
              – Ed999
              Jan 23 at 18:23











              0












              $begingroup$

              When An object is on a cliff at rest then relative to ground ,it will have 0 kinetic energy.we can take the surface of the cliff at 0 potential. When it is on surface of cliff,Two forces are working on it, The gravitational force of earth and the normal force of surface.Net force is zero, So it is in equilibrium.As net force is zero,so change in it's kinetic energy or its potential energy is also zero ($F=-dU/dx$).



              When it is falling from the cliff ,we removed the normal force ,so net force is acting that is gravitational force. This will change its potential (kinetic)energy. We assumed the cliff at 0 potential energy ,so the force will decrease the potential energy in negative as wolframe johny said.



              The “energy” is just an abstract accounting tool that we use to describe the results of those interactions in terms of the work being done.



              We talk about the potential energy of system ,because we need force that will cause change in potential energy,but kinetic energy is just because of the relative motion of the object






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                When An object is on a cliff at rest then relative to ground ,it will have 0 kinetic energy.we can take the surface of the cliff at 0 potential. When it is on surface of cliff,Two forces are working on it, The gravitational force of earth and the normal force of surface.Net force is zero, So it is in equilibrium.As net force is zero,so change in it's kinetic energy or its potential energy is also zero ($F=-dU/dx$).



                When it is falling from the cliff ,we removed the normal force ,so net force is acting that is gravitational force. This will change its potential (kinetic)energy. We assumed the cliff at 0 potential energy ,so the force will decrease the potential energy in negative as wolframe johny said.



                The “energy” is just an abstract accounting tool that we use to describe the results of those interactions in terms of the work being done.



                We talk about the potential energy of system ,because we need force that will cause change in potential energy,but kinetic energy is just because of the relative motion of the object






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  When An object is on a cliff at rest then relative to ground ,it will have 0 kinetic energy.we can take the surface of the cliff at 0 potential. When it is on surface of cliff,Two forces are working on it, The gravitational force of earth and the normal force of surface.Net force is zero, So it is in equilibrium.As net force is zero,so change in it's kinetic energy or its potential energy is also zero ($F=-dU/dx$).



                  When it is falling from the cliff ,we removed the normal force ,so net force is acting that is gravitational force. This will change its potential (kinetic)energy. We assumed the cliff at 0 potential energy ,so the force will decrease the potential energy in negative as wolframe johny said.



                  The “energy” is just an abstract accounting tool that we use to describe the results of those interactions in terms of the work being done.



                  We talk about the potential energy of system ,because we need force that will cause change in potential energy,but kinetic energy is just because of the relative motion of the object






                  share|cite|improve this answer









                  $endgroup$



                  When An object is on a cliff at rest then relative to ground ,it will have 0 kinetic energy.we can take the surface of the cliff at 0 potential. When it is on surface of cliff,Two forces are working on it, The gravitational force of earth and the normal force of surface.Net force is zero, So it is in equilibrium.As net force is zero,so change in it's kinetic energy or its potential energy is also zero ($F=-dU/dx$).



                  When it is falling from the cliff ,we removed the normal force ,so net force is acting that is gravitational force. This will change its potential (kinetic)energy. We assumed the cliff at 0 potential energy ,so the force will decrease the potential energy in negative as wolframe johny said.



                  The “energy” is just an abstract accounting tool that we use to describe the results of those interactions in terms of the work being done.



                  We talk about the potential energy of system ,because we need force that will cause change in potential energy,but kinetic energy is just because of the relative motion of the object







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 23 at 16:38









                  TheBrolyTheBroly

                  1




                  1























                      0












                      $begingroup$

                      If the ball was on top of a cliff, then it already had gravitational potential energy. This comes from the fact that if the ball fell off the cliff, if would gain kinetic energy (while losing the potential energy it had from being on top of the cliff).



                      Gravitational potential energy is usually discussed with respect to a system because there needs to be other mass(es) exerting gravity on an object to give it said potential energy. However, there are other types of potential energy internal to an object (e.g., chemical potential energy), that has no reliance on an external system.



                      The main thing to remember is that potential energy is how much kinetic energy an object could potentially exert. An object being on top of a cliff has potential energy since it has the potential to drop off the cliff and gain kinetic energy.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I think there's something missing here about the arbitrariness of the zero point of kinetic/potential energy. Whatever numerical value we choose to assign to the potential energy of a ball at the top of a cliff is rather meaningless. Suppose the ball falls to the bottom of the cliff, gaining a certain amount of KE before it lands. Now it turns out the ball landed on a trap door and falls through, gaining even more KE, even though in the frame of reference we started in, the ball didn't ever have that much PE. All that matters is the change in KE and PE, but not the actual numerical values.
                        $endgroup$
                        – Nuclear Wang
                        Jan 23 at 20:39
















                      0












                      $begingroup$

                      If the ball was on top of a cliff, then it already had gravitational potential energy. This comes from the fact that if the ball fell off the cliff, if would gain kinetic energy (while losing the potential energy it had from being on top of the cliff).



                      Gravitational potential energy is usually discussed with respect to a system because there needs to be other mass(es) exerting gravity on an object to give it said potential energy. However, there are other types of potential energy internal to an object (e.g., chemical potential energy), that has no reliance on an external system.



                      The main thing to remember is that potential energy is how much kinetic energy an object could potentially exert. An object being on top of a cliff has potential energy since it has the potential to drop off the cliff and gain kinetic energy.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        I think there's something missing here about the arbitrariness of the zero point of kinetic/potential energy. Whatever numerical value we choose to assign to the potential energy of a ball at the top of a cliff is rather meaningless. Suppose the ball falls to the bottom of the cliff, gaining a certain amount of KE before it lands. Now it turns out the ball landed on a trap door and falls through, gaining even more KE, even though in the frame of reference we started in, the ball didn't ever have that much PE. All that matters is the change in KE and PE, but not the actual numerical values.
                        $endgroup$
                        – Nuclear Wang
                        Jan 23 at 20:39














                      0












                      0








                      0





                      $begingroup$

                      If the ball was on top of a cliff, then it already had gravitational potential energy. This comes from the fact that if the ball fell off the cliff, if would gain kinetic energy (while losing the potential energy it had from being on top of the cliff).



                      Gravitational potential energy is usually discussed with respect to a system because there needs to be other mass(es) exerting gravity on an object to give it said potential energy. However, there are other types of potential energy internal to an object (e.g., chemical potential energy), that has no reliance on an external system.



                      The main thing to remember is that potential energy is how much kinetic energy an object could potentially exert. An object being on top of a cliff has potential energy since it has the potential to drop off the cliff and gain kinetic energy.






                      share|cite|improve this answer









                      $endgroup$



                      If the ball was on top of a cliff, then it already had gravitational potential energy. This comes from the fact that if the ball fell off the cliff, if would gain kinetic energy (while losing the potential energy it had from being on top of the cliff).



                      Gravitational potential energy is usually discussed with respect to a system because there needs to be other mass(es) exerting gravity on an object to give it said potential energy. However, there are other types of potential energy internal to an object (e.g., chemical potential energy), that has no reliance on an external system.



                      The main thing to remember is that potential energy is how much kinetic energy an object could potentially exert. An object being on top of a cliff has potential energy since it has the potential to drop off the cliff and gain kinetic energy.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 23 at 19:05









                      Inertial IgnoranceInertial Ignorance

                      501119




                      501119












                      • $begingroup$
                        I think there's something missing here about the arbitrariness of the zero point of kinetic/potential energy. Whatever numerical value we choose to assign to the potential energy of a ball at the top of a cliff is rather meaningless. Suppose the ball falls to the bottom of the cliff, gaining a certain amount of KE before it lands. Now it turns out the ball landed on a trap door and falls through, gaining even more KE, even though in the frame of reference we started in, the ball didn't ever have that much PE. All that matters is the change in KE and PE, but not the actual numerical values.
                        $endgroup$
                        – Nuclear Wang
                        Jan 23 at 20:39


















                      • $begingroup$
                        I think there's something missing here about the arbitrariness of the zero point of kinetic/potential energy. Whatever numerical value we choose to assign to the potential energy of a ball at the top of a cliff is rather meaningless. Suppose the ball falls to the bottom of the cliff, gaining a certain amount of KE before it lands. Now it turns out the ball landed on a trap door and falls through, gaining even more KE, even though in the frame of reference we started in, the ball didn't ever have that much PE. All that matters is the change in KE and PE, but not the actual numerical values.
                        $endgroup$
                        – Nuclear Wang
                        Jan 23 at 20:39
















                      $begingroup$
                      I think there's something missing here about the arbitrariness of the zero point of kinetic/potential energy. Whatever numerical value we choose to assign to the potential energy of a ball at the top of a cliff is rather meaningless. Suppose the ball falls to the bottom of the cliff, gaining a certain amount of KE before it lands. Now it turns out the ball landed on a trap door and falls through, gaining even more KE, even though in the frame of reference we started in, the ball didn't ever have that much PE. All that matters is the change in KE and PE, but not the actual numerical values.
                      $endgroup$
                      – Nuclear Wang
                      Jan 23 at 20:39




                      $begingroup$
                      I think there's something missing here about the arbitrariness of the zero point of kinetic/potential energy. Whatever numerical value we choose to assign to the potential energy of a ball at the top of a cliff is rather meaningless. Suppose the ball falls to the bottom of the cliff, gaining a certain amount of KE before it lands. Now it turns out the ball landed on a trap door and falls through, gaining even more KE, even though in the frame of reference we started in, the ball didn't ever have that much PE. All that matters is the change in KE and PE, but not the actual numerical values.
                      $endgroup$
                      – Nuclear Wang
                      Jan 23 at 20:39


















                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Physics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f456150%2fhow-can-an-object-with-zero-potential-and-kinetic-energy-ever-move%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      -

                      Popular posts from this blog

                      How to change which sound is reproduced for terminal bell?

                      Image
                      3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
                      Read more

                      Can I use Tabulator js library in my java Spring + Thymeleaf project?

                      Image
                      0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
                      Read more

                      Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                      Image
                      up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
                      Read more