An Energy Conservation Puzzle
Posted on November 12, 2007
Filed Under Physics Puzzles |
Here’s a puzzle that makes lots of experienced physicists pause for a moment - or two!
Suppose you have a cart at the top of an inclined hill. Initially, the cart only has potential energy, equal to mgh. As it descends to the bottom of the mountain, its potential energy is converted into kinetic energy, and we can even calculate its magnitude using the principle of conservation of energy: Kinetic = Potential, or:
The whole process is illustrated in a drawing I’ve made:
So far so good - nothing radical has been said. This is a standard problem. Consider, however, what happens if we now view the process from another (inertial, mind you!) frame of reference moving with a velocity equal to the final velocity of the cart, 
. Think for a moment about the cart’s initial and final states in this new frame of reference. Initially, due to the frame’s velocity, the cart seems to move “backwards” at the top of the hill, and when it reaches the bottom it seems to be stationary. But wait, all is not well in the kingdom of Denmark! It seems that energy conservation has suffered a severe blow, since now the cart has non-zero energy at the beginning (kinetic + potential), and zero energy at the bottom of the hill (no kinetic, no potential). This is illustrated below:
This begs the questions:
- Has conservation of energy failed? How is that possible? How can energy be conserved in one frame and not another? Note that both frames are inertial, so you can’t argue there are fictitious forces involved.
- If it hasn’t failed, where did the energy go?
Enjoy the riddle! 
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17 Responses to “An Energy Conservation Puzzle”
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But… the v in the second one is negative so you end up with 0=0
I have a test on conservation of energy stuff, coming up this week.
Yes, you are correct, but the kinetic energy contains the velocity SQUARED, making it positive - so it’s back to the drawing board.
You need to take into account the velocity of the cart in the new inertial frame — ie. shifted by v.
I’m not sure I follow you. The velocity in the new frame has been taken into account already. It is the reason the cart appears moving at the top of the slope, and that it appears at rest at the bottom.
The cart is running uphill in the second frame. Or rather the accelerating force has apparently swapped symmetry. g is negative
Momentum is not ficticious. Force = change in mv. v is opposite in second frame.
Hi Mark,
As noted before, both frames of reference are inertial. There are no fictitious forces involved. In particular, g cannot change sign and is the same magnitude and direction as always.
Regarding your second post, think of two (inertial) observers analyzing the same experiment and presenting seemingly two different conclusions. This eliminates the conceptual need to “switch from one frame to the other”, so there is no “change in mv”.
I admit bein’ rusty. I’m retired after 31 years underground coal mining. I did get a BA in philosophy before I started working. But now I feel dumber than owlshit. Have patience.
Inertial frames honor only first two laws of motion. So a physical force of some kind has to act on the cart to accelerate it. The second law works when observed from an inertial frame only. The inertial frame is defined by the first law. And since the equal but opposite forces of the third law are not allowed, is it any wonder that the Kingdom of Denmark can not see a conservation of energy?
OK that’s bluff #1 Love this stuff by the way.
well i think you’ve calculated different things in different places.let’s assume the stationary frame is X and the moving is X’,and
delta(h)=height of hill=top of hill-bottom of hill=h(regardless of X or X’]
from the point of view of X:
E(initial) cart=1/2 m.v^2+m.g.delta(h) [v=speed of X’ relative to X]
E(final) cart=1/2 m.v^2+1/2 m.v^2=m.v^2=E(init)
from the point of view of X’:
E’(initial) cart=m.g.delta(h)
E’(final) cart=1/2 m.v^2=E’(init)
[v=speed of cart relative to X’]
OK. I’m gettin’ there. One of the two frames is not valid. I gotta work on a car tomorrow and it’s 2:30 in the morning here. More tomorrow.
Answer 1: The conservation of energy is one of the basic laws of the nature. It cannot be violated anywhere.
Answer 2: Work done by the Normal Reaction.
Intuitively:
- In frame 1, as the cart rolls down its speed increases, which matches to the direction of [g], hence [g] is positive.
- In frame 2, as the cart rolls down its speed decreases to a halt, which means the [g] in this case is negative.
The conservation of energy is, of course, not violated (in frame 2 its Ei=Ef=zero)
But my intuition has been known to fail in physics… so I need to find a more formulated answer. I have a hard time imaging the ground moving and the the cart starting from speed and slowing down…
1a. Yes. 1b. In the second frame the first law fails. The observer there notes not only a change in the carts velocity, but a second observed change in the carts velocity relative to the slope and surrounding scenery. So which is doing the accelerating? The cart or the earth it is on? That is why 2nd frame fails. Because the formulas for energy conservation for an additional frame is not being accounted for. The one in the 2nd frame where the scenery is seen as accelerating relative to the cart or vice versa. Since we know the force on the cart is the same in both frames. Only in the first do we see a non failure of the first law.
2. is moot.
Ok that’s bluff #2. And I am glad that damn cart wasn’t also on a turntable.
I believe Himanshu is correct. In inertial frames, we have the work energy theorem:
Change in kinetic energy of some object = Work done by all forces on the same object
For simplicity assume the slope is a flat inclined plane making some angle theta with the horizontal. In the first frame where the slope is at rest, the reason why we can ignore the normal force is because it is perpendicular to the motion, i.e. F[normal].dx = 0. But in the second frame where the slope is in motion, a simple sketch will tell us that at all time while the “normal” force acts in the direction (Pi/2 - theta) from the upward vertical and the gravitational force act in downward vertical direction, the motion of the cart follows a curved path in space. Hence neither force is perpendicular to the motion at all times and must be included for computation of the work done.
The “normal” force is responsible for removing the energy from the cart in the second frame.
It is likely that one can come up with a similarly interesting puzzle with electric and magnetic fields. In a given frame, magnetic fields do not do work, since it contributes a force of v x B, i.e. perpendicular to the motion. However, upon changing from one inertial frames to another magnetic fields transform into part magnetic plus electric — it is the transformation epsilon_{ijk} B_k = F_{ij} -> F_{mu nu} Lambda^mu_i Lambda^nu_j — and in the new frame it would contribute to work done on an electrically charged particle.
Hi,
okay this is quite radical and people will start snickering when they hear this(and I have a deficiency in phrasing anything beautifully), BUT…
Is the weight vector facing upwards when the cart is uphill…due to the fact that the cart is actually upside down and that the earth or ground starts on top of that black squiggly line, and air/atmosphere/space on the bottom?
If that is true, velocity must be zero on bottom of the hill (or on top of the rollercoaster ramp scenario I have played out in my mind).
okay, now that i re-read the question, I’m getting dissappointed by my answer…I’m going to attempt this problem later, when it’s not 2 a.m.
btw love the puzzles! I wish i hadn’t drained my mind this summer before I start freshman year…
Hi Scribed,
Nope, the weight vector points downwards towards the Earth at all times. I didn’t understand your comment about the cart being upside-down.
Assaf.
PS
Good luck with your studies!
oh, I meant the diagram flipped over, but i guess not…
gonna have a look later