# The Vanishing Energy

zorro
Product of my imagination (not a homework/textbook question)-

A body is kept on top of a hill of height 'h'. If we fix the 0 potential energy at the ground, the body has an initial energy - mgh. Assume that the hill surface + ground is frictionless. When the body starts sliding (with negligible push), it gains a velocity of v=(2gh)1/2 on reaching the ground, which follows from energy conservation.

Now suppose we observe the whole situation from a reference frame moving with a uniform velocity v=(2gh)1/2 on the ground. Initially, the body has a total energy given by 0.5mv2 + mgh. On reaching the bottom, the total energy is 0!

Where does the energy vanish?

## Answers and Replies

Homework Helper
Hi Abdul!

Interesting!!

When the hill is stationary, the normal reaction force is (obviously) parallel to the displacement , and so does no work.

But when the hill is moving with speed √(2gh), the normal reaction force is not parallel to the displacement, and so the work done is equal to … erm

ooh! … your problem … you do the maths!

xlines
Hi Abdul!

Interesting!!

When the hill is stationary, the normal reaction force is (obviously) parallel to the displacement , and so does no work.

But when the hill is moving with speed √(2gh), the normal reaction force is not parallel to the displacement, and so the work done is equal to … erm

ooh! … your problem … you do the maths!

perpendicular?

zorro
the normal reaction force is not

Yeahh!! You are right. The work done by the ground = change in KE of the block = -0.5mv2

So work done by the block = 0.5mv2

Thanks!!

Homework Helper
Where does the energy vanish?
Into the earth. If you include the effect on the earth, then total momentum (angular and linear) and total energy (angular and linear) of the earth + body system remains constant (assuming no external forces or losses (like heat)), as observed from any inertial (non-accelerating) frame of reference. Take a look at post #3 in this thread:

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Gold Member
Hi Abdul!

Interesting!!

When the hill is stationary, the normal reaction force is (obviously) parallel to the displacement , and so does no work.

But when the hill is moving with speed √(2gh), the normal reaction force is not parallel to the displacement, and so the work done is equal to … erm

ooh! … your problem … you do the maths!

I don't believe this solves the problem because we could remove the hill entirely and be left with basically the same problem.

The important thing to realize is that conservation of energy and conversation of momentum are closely connected when switching frames. It is conservation of momentum that guarantees conservation of energy holds when you shift frames of reference. It becomes extremely important that you keep track of what's happening to the earth if it is moving.

The reason you don't normally need to think about the earth moving, but you do in this case, is the nonlinear nature of kinetic energy. The kinetic energy is:
$$\frac{1}{2}mv^{2}$$
Conservation of momentum tells us that if the earth is stationary and a small body changes its velocity by $\Delta v$, then
$$m(v+\Delta v) + Mv_{e}=mv \rightarrow v_{e}=-\frac{m}{M}\Delta v \equiv -\delta \Delta v$$
(M=mass of earth, m=mass of object, v_e=final velocity of earth, $\delta$=m/M)
The kinetic energy of the earth is then given by
$$\frac{1}{2}Mv_{e}^2 = \frac{1}{2}M\delta^2(\Delta v)^2 = \frac{1}{2}m\delta(\Delta v)^2$$
assuming that the velocity of the earth is 0 initially. This quantity is negligibly small because of the small parameter $\delta$ that has no large factor like the mass of the earth to counter its effect. Now assume, as in your example that the earth starts with velocity vi, and look at the change in kinetic energy of the earth. Of course the change in velocity of the earth will be the same when we shift reference frames, so lets still call that change ve, which will have the same value as before:
$$\Delta KE = \frac{1}{2}M(v_{i} + v_{e})^2 - \frac{1}{2}Mv_{i}^2=\frac{1}{2}Mv_{i}^2 + Mv_{e}v_{i} + \frac{1}{2}Mv_{e}^2 - \frac{1}{2}Mv_{i}^2 = Mv_{e}v_{i}$$
Now plug in the value of ve from conservation of momentum:
$$\Delta KE = Mv_{i}\delta v_{i} + \frac{1}{2}M\delta^2v_{e}^2= mv_{i}^2 + \frac{1}{2}m\delta v_{e}^2\approx 2(\frac{1}{2}mv_{i}^2)= mgh + \frac{1}{2}mv_{i}^2$$
I again neglected the small term with a $\delta$ still in it. In the last equality I assumed we were working with your example where the initial velocity was chosen so that the kinetic energy of the small mass was the same as the potential energy. As you see, this is exactly the energy you started with in the shifted frame, and which seemed to go to 0 when you looked only at the small mass. Now it is clear what happened: in this frame the energy gets transfered to the earth.

The main point is that you can't neglect the change in kinetic energy of a very large body when it doesn't start from rest. Since kinetic energy is proportional to the square of velocity, and the derivative of kinetic energy is therefore proportional to velocity, changes in kinetic energy of the large body will be negligibly small when they change from rest, but the same change in v will produce a considerable change in kinetic energy when it does not.

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Homework Helper
I don't believe this solves the problem because we could remove the hill entirely and be left with basically the same problem.

No!

Remove the hill, and you have a free-falling body.

Its horizontal component of velocity stays the same, so there is no frame in which that component can be reduced.

The whole point of Abdul Qadeer's problem was that he constructed it so that a change in frame could reduce the horizontal component!
The important thing to realize is that conservation of energy and conversation of momentum are closely connected when switching frames. It is conservation of momentum that guarantees conservation of energy holds when you shift frames of reference.

No, they both derive from the translational symmetry of Galilean space-time (a simple case of Noether's theorem).
It becomes extremely important that you keep track of what's happening to the earth if it is moving.

The reason you don't normally need to think about the earth moving, but you do in this case, is the nonlinear nature of kinetic energy.

Now it is clear what happened: in this frame the energy gets transfered to the earth.

You can play this "let's consider the Earth" game in almost any situation. Obviously, most experiments are connected to Earth in some way, and if so you can always deliberately add the Earth to the system you're considering.

But that doesn't mean that the system without the Earth is invalid, or is an approximation.

(As approximations go, btw, the energy anomaly in Abdul's problem is huge, and cannot possibly be put down to an error in approximation)

Newton's laws apply to a body on its own, whether it is in contact with other bodies or not

(this is so obvious it needs neither explanation nor example)

they apply to the body sliding down the hill on its own

all you have to do is to add the correct forces, and "turn the handle"​

In this case, one of those forces happens to be doing work in the alternative frame …

end of

Gold Member
No!

Remove the hill, and you have a free-falling body.

Its horizontal component of velocity stays the same, so there is no frame in which that component can be reduced.

The whole point of Abdul Qadeer's problem was that he constructed it so that a change in frame could reduce the horizontal component!

I thought of the hill as vertical at the bottom, but horizontal would make more sense as an assumption, although he doesn't say.

The problem with your analysis is that it doesn't explicitly show where the energy went. Yes the hill does work, but the energy balance at the end is exactly AbdulQadeer's if you don't consider the energy of the earth.

Of course the solution is that while the hill does work on the body, the body does work on the earth, so the energy goes into the earth.

Newton's laws apply to a body on its own, whether it is in contact with other bodies or not
True, but conservation of energy obviously does not. Moving to an open system is one answer to the question because you now have an external work term that does what its supposed to in the case of a horizontal bottom. It doesn't show where the energy goes in the universe, though.
But that doesn't mean that the system without the Earth is invalid, or is an approximation.

(As approximations go, btw, the energy anomaly in Abdul's problem is huge, and cannot possibly be put down to an error in approximation)
My point was that it was a valid approximation in the case of a stationary earth but not a moving earth as far as conservation of energy is concerned.
When you drop an apple, the apple falls and the earth (center of mass of the planet not including the apple) rises to meet it. There is kinetic energy added to both bodies when you look from the frame where the earth is initially stationary. Ignoring one kinetic energy in the energy balance is an approximation, but a good one.
If you go to a moving frame, my previous post shows why this becomes a bad approximation. It's a bit of semantics whether the broken energy balance results from an approximation since it is such a bad approximation that it probably doesn't deserve to be called an approximation at all at that point. The change in velocity is tiny, but the mass of the earth is enormous. When you get the cross term in kinetic energy where the velocity of the earth is to the first power, the massiveness of the earth can counter the tininess of the velocity to give an important term.

edit: Just realized the the velocity coming off the hill can't be totally downward since if there is any horizontal component at all the mass would slip off the hill and not loose the horizontal component, so a horizontal bottom is definitely what he meant.

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