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## Homework Statement

A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m/s. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact ? Would your answer be different if the elevator were stationary ?

The problem I am facing is that the answer in book is given as 8.82 J and it mentions that it will be the same even when lift is stationary. The answer that I get is 25.5 J as explained below. May be I am missing something here.

## Homework Equations

I am assuming that the observer is on ground and not in elevator i.e. frame of reference is ground.

Also, g is assumed to be 10 m/s

^{2}.

KE1 + PE1 = KE2 + PE2, where 1 refers to when bolt starts its fall and 2 refers to just before it hits the floor.

Also for the bolt fall,

v = u + at and v

^{2}= u

^{2}+ 2 x a x s

## The Attempt at a Solution

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The bolt falls from a height with initial velocity of 7 m/s since bolt will have same velocity as the elevator when it starts it's fall. When it hits the floor it has fallen a distance of (3 + 7t) since floor of elevator descends by 7t

So, if the floor of the elevator when bolt hits the floor is taken as datum level for potential energy, then

KE1 + PE1 = KE2 + PE2, where 1 refers to when bolt starts its fall and 2 refers to just before it hits the floor.

.KE2 + PE2 = .5 x .3 x v

^{2}+ 0

Also, using kinematics, we have

v = 7 + 10 x t and v

^{2}= 7

^{2}+ 2 x 10 x (3 + 7t)

Solving above equations, t = .78 seconds, v= 14.8 m/s

Since bolt does not rebound after hitting the floor, so KE2 + PE2 - (KE of bolt after hitting floor) = Heat generated. = .5 x .3 x 14.8

^{2}+ 0 - .5 x .3 x 7

^{2}= 25.5 J.

But answer at back of book is given as 8.82 J. I am not getting where I am making a mistake.

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