A block of mass 2 kg starts from rest at x=0 on a horizontal surface. It is pulled by a constant external force of 20 N in the positive x -direction. The frictional force f acting on the block varies with position x such that it decreases linearly from 20 N at x=0 to 0 N at x=4 m . The kinetic energy of the block at x=4 m is :

Step-by-step solution

The work done by the constant external force is: W_ ext = F d = 20 4 = 80 J The frictional force decreases linearly with position. The work done by friction is the negative of the area under the force-position graph (which forms a right-angled triangle): W_ friction = - ( 1 2 base height ) W_ friction = - ( 1 2 4 20 ) = -40 J The net work done on the block is: W_ net = W_ ext + W_ friction = 80 - 40 = 40 J By the work-energy theorem, the net work done is equal to the change in kinetic energy: W_ net = K_f - K_i Sin