Question

A block slides with constant velocity on a plane inclined at an angle theta. The same block is pushed up the plane with an initial velocity v''. The distance covered by the block before coming to rest is

Answer 1

Answer:

As the block is sliding down with constant velocity. Therefore friction force along the slope must be equal to the component of gravity along the slope.

∴μN=mgsinθ⇒μmgcosθ=mgsinθ⇒μ=tanθ

When thrown up:

Initial Velocity is Vo

Acceleration (both friction and component of gravity acting downwards the slope as relative motion is upwards) is μN+mgsinθ=tanθmgcosθ+mgsinθ=2mgsinθ

Final Velocity is zero. 

Therefore distance traveled up the incline is:

v2−u2=2as⇒0−Vo2=2(−2mgsinθ)s⇒s=4mgsinθVo2

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