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
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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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