A 100 kg block is started with a speed of `2.0 m s^(-1)` 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 sutuation from a frame of reference moving at `2.0 m s^(-1)` 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 starts moving with the belt at `2.0 m s^(-1)`, 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.

Answer

A, B, C, D

Explanation

Mass of block =`100kg.,u=2m/s`
`u=0*2,v=0`
`We Know `dQ=du+dw`
`In this case`dQ=du+ dw`
` In this case `dQ=0`
`rArr`-du=dw`
`rArr` du=-((1)/(2)mv^2-(1)/(2)mv^2)`
`=`-(1)/(2)xx100xx2xx2`
`=200J`