Q.90. Work/Energy, Inclined plane.

Question 90.
The top of an inclined plane is at a height of 0.7m above the bottom. A block of mass 0.2kg is released from rest at the top of the plane and slides a distance of 2.5m to the bottom. Find the kinetic energy of the block when it reaches the bottom of the plane in each of the following cases:
(i) the plane is smooth,
(ii) the coefficient of friction between the plane and the block is 0.15.

Answer 90.
( i )
P.E. at the top is converted into K.E. at the bottom of the plane.
=> (0.5) mv^2 = mgh
=> v = √(2gh) = √(2 * 9.8 * 0.7)
= 3.7 m/s.

( ii )
Net force downwards
= mg cosθ - μmg sinθ, where
θ = angle made by the inclined plane with the vertical
cosθ = 0.7/2.5 = 0.28
=> sinθ = 0.96

Work done = Force x displacement
= mg [ cosθ - μ sinθ ] * (2.5) J.
K.E. = work done
=> (0.5) mv^2 = mg [ cosθ - μ sinθ ] * (2.5)
=> v = √ [2g [ cosθ - μ sinθ ] * (2.5)]
=> v = √ [ 19.6 [0.28 - (0.15) * (.96)] * (2.5)
=> v = 2.58 m/s.

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