Consider an elevator moving down with uniform velocity. A person standing inside watches an object fall from the ceiling of the elevator to the floor. Say the height of the elevator is h. Then the work done by gravity in that frame of reference should be mgh. But consider this same event being watched by someone else in the stationary frame of reference. In his reference frame, the object travels a larger distance as it falls from the ceiling to the floor of the elevator because the floor itself is moving downwards (one can calculate this extra distance covered to be u sqrt(2h/g)) and hence the change in kinetic energy should be more in that frame than in the moving frame. Shouldn't the change in kinetic energy be more in a moving elevator from a stationary frame of reference?

djo57bgfrqn

djo57bgfrqn

Answered question

2022-10-20

Consider an elevator moving down with uniform velocity. A person standing inside watches an object fall from the ceiling of the elevator to the floor. Say the height of the elevator is h. Then the work done by gravity in that frame of reference should be m g h. But consider this same event being watched by someone else in the stationary frame of reference. In his reference frame, the object travels a larger distance as it falls from the ceiling to the floor of the elevator because the floor itself is moving downwards (one can calculate this extra distance covered to be u 2 h g ) and hence the change in kinetic energy should be more in that frame than in the moving frame.
Shouldn't the change in kinetic energy be more in a moving elevator from a stationary frame of reference?

Answer & Explanation

getrdone07tl

getrdone07tl

Beginner2022-10-21Added 23 answers

See, you are assuming that, after the collision, the velocity of the ball-elevator ensemble is u, but this is not fully true: it will be u = u + m m + M 2 g h , M being the mass of the elevator. Of course if M that reduces to u = u, but when computing the KE, something funny happens:
1 2 ( m + M ) u 2 = 1 2 ( m + M ) u 2 + m 2 m + M g h + u m 2 g h
That last term which does not depend on M is the key here. Of course the first term, with the ( m + M ) dominates the others, but it will be cancelled out by identical terms in the KE before the collision. But if you assume that because M you can take u = u, you will be missing this last term, which exactly cancels out that extra energy.
Doing the math for a finite elevator mass, and using conservation of momentum to compute the final velocity, you eventually get to energy lost in an inellastic collision to be 1 2 m M m + M ( u v ) 2 , which for M reduces to 1 2 m ( u v ) 2 .
Aydin Jarvis

Aydin Jarvis

Beginner2022-10-22Added 1 answers

There is no such thing as conservation of energy between intertial reference frames.
Considering the observer inside the elevation, the free fall takes t f = 2 h g , after which is has velocity g t f , and thus kinetic energy m g h (which is cheating, as this is what you used to calculate t f in the first place. However, the total energy of the particle itself is conserved within this frame, between two times.
Now consider the external observer. It sees an increase in kinetic energy of
Δ K = 1 2 m ( u + 2 g h ) 2 1 2 m u 2
Which simplifies to:
Δ K = 1 2 m ( u 2 + 2 u 2 g h + 2 g h ) 1 2 m u 2
= m u 2 g h + m g h
Where the first term is related precisely to the additional difference in height that you calculated.

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