A solid spherical ball rolls on a horizontal surface at 10 m s - 1 and continues to roll up on an inclined surface as shown in the figure. If the mass of the ball is 11 kg and frictional losses are negligible, the value of h , where the ball stops and starts rolling down the inclination is (Assume g = 10 m s - 2 )

Question

A solid spherical ball rolls on a horizontal surface at 10 m s - 1 and continues to roll up on an inclined surface as shown in the figure. If the mass of the ball is 11 kg and frictional losses are negligible, the value of h , where the ball stops and starts rolling down the inclination is (Assume g = 10 m s - 2 ), 8 m, 6 m, 7 m, 10 m

MARKS App · Accepted Answer

Given, Initial velocity of solid spherical ball is v = 10 m s - 1 Mass of ball rolling on the surface is m = 11 kg Applying work energy theorem, W = ∆ K E W g = K E f - K E i . . . 1 Now, initial kinetic energy of spherical ball is K i = 1 2 m v 2 1 + k 2 R 2 And final kinetic energy is given by K f = 0 Work done due to gravity is W g = - m g h Substitute the values in equation (1) we get ⇒ - m g h = 0 - 1 2 × 11 × 10 2 1 + 2 5 Where, k 2 R 2 = 2 5 ⇒ h = 350 50 = 7 m

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