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A 100 kg block is started with a speed of 2`0 m/s on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0`20. (a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt, (b) Consider the situation from a frame of reference moving at 20 m/s along the initial velocity of the block. As seen from this frame, the block is gently put on a moving belt and in due time the block startsmoving with the belt at 2`0 m/s. Calculate the increase in the kinetic energy of the block as it stops slipping past the belt, (c) Find the work done in this frame by the external force holding the belt.

A 100 kg block is started with a speed of 2`0 m/s on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0`20. (a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt, (b) Consider the situation from a frame of reference moving at 20 m/s along the initial velocity of the block. As seen from this frame, the block is gently put on a moving belt and in due time the block startsmoving with the belt at 2`0 m/s. Calculate the increase in the kinetic energy of the block as it stops slipping past the belt, (c) Find the work done in this frame by the external force holding the belt.

Grade:12th pass

2 Answers

Arun
25750 Points
6 years ago
Dear student
 
 mass of block = 100 kg u = 2 m/s, m = 0.2 v = 0 dQ = du + dw In this case dQ = 0 ⇒ - du = dw ⇒ du = - (- ½ mv^2 – ½ mu^2) = ½ x 100x 2x 2 = 200 J   RegardsArun (askIITians forum expert)
RAUSHAN KUMAR
10 Points
6 years ago
Sir plz gives the solution of (c) Find the work done in this frame by the external force holding the belt

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