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A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with a uniform speed of 7 ms-1. It hits the floor of the elevator (length of 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 ?

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Since the motion is unifirm speed motion, the answer will not be whether the elevator is stationary or moving, as in either case the relative motion's dynamics remains the same. so the total energy gained before it touches the floor of the elevator is mgh=.3*9.8*3=8.82 j , assuming there was no loss...
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Since the motion is unifirm speed motion, the answer will not be whether the elevator is stationary or moving, as in either case the relative motion's dynamics remains the same.

so the total energy gained before it touches the floor of the elevator is mgh=.3*9.8*3=8.82 j , assuming there was no loss of energy in the form of sound or due to friction of air while the bolt was falling.

assumption: g=9.8 which will be true near surface of earth.

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JEE Main Maths, Physics With 12 years exp.

Heat produced=loss of potential energy=mgh=0.3×9.8×3=8.82J. The answer will be same even if the elevator is stationary. Since it was moving with constant velocity
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By using law of conservation of energy heat produced by the impact will be equal to loss in potential energy.H = mghm = 0.3kg g=9.8m/s² h=3mHeat produced = 0.3*9.8*3 = 8.82 JHeat produced is the same for all cases as in each and every case relative velocity of bolt is zero.
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By using law of conservation of energy heat produced by the impact will be equal to loss in potential energy.

H = mgh

m = 0.3kg g=9.8m/s² h=3m

Heat produced = 0.3*9.8*3 = 8.82 J

Heat produced is the same for all cases as in each and every case relative velocity of bolt is zero.

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