First slide
Friction
Question

A block of mass m is projected up an inclined plane of inclination θ with an initial velocity u. If the coefficient of kinetic friction between the block and the plane is μ = tan θ, the distance up to which the block will rise up the plane, before coming to rest, is given by

Moderate
Solution

Refer to Fig. Since the block is projected upwards, it rises after overcoming two forces: (i) the component mg sin θ of the weight mg and (ii) the force of friction F= mg sin θ, both acting downwards. Therefore, the total downward acceleration is

a=gsinθgsinθ=2gsinθ

Lets be the distance moved up the plane before the block comes to rest. Then, from v2u2=2as, we have

0u2=2×(2gsinθ)×s

   or   s=u24gsinθ

Hence, the correct choice is (3).

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