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A stationary body of mass m is slowly lo...

A stationary body of mass m is slowly lowered on to a massive platform `(M gt gt m)` moving at a speed `v_(0) = 4 m//s` (Fig. ) . How long will the body slide along the platform and what distance will it travel during this time? The coefficient of friction is `mu=0.2`.

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The correct Answer is:
2s ; 4m
Let us go over to the reference frame connected with the platform. Here the initial speed to the body is `-v_(0)` and the final is zero. We have
`a=(T)/(m) = (mu mg)/(m) = mu g, tau = (0-(-v_(0)))/(a)=(v_(0))/(mug)`
`x= -v_(0) tau +(a tau^(2))/(2) = - (v_(0)^(2))/(2mu g)`
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