A block of mass 2 kg is pulled along a rough horizontal surface by an external force. The velocity of the block varies with its position x as v = 3x , where v is in m/s and x is in m. If the coefficient of kinetic friction between the block and the surface is 0.5 , the work done by the external force during the displacement of the block from x = 0 m to x = 2 m is (Take g = 10 m/s ^2 ):

Step-by-step solution

The initial velocity of the block at x = 0 m is v_i = 3(0) = 0 m/s. The final velocity at x = 2 m is v_f = 3(2) = 6 m/s. The change in kinetic energy of the block is: K = 1 2 m (v_f^2 - v_i^2) = 1 2 2 (6^2 - 0) = 36 J. The work done by the kinetic friction force is: W_ fric = - _k m g d = -0.5 2 10 2 = -20 J. According to the work-energy theorem, the net work done on the block equals its change in kinetic energy: W_ ext + W_ fric = K W_ ext - 20 = 36 W_ ext = 56 J. Answer: 56 J