A block of mass m is pulled along a horizontal rough surface with a constant coefficient of kinetic friction . A downward vertical force that varies with position as F_y = cx (where c is a positive constant) is applied to the block as it moves from x=0 to x=d . The magnitude of the total work done by the frictional force on the block is :

Step-by-step solution

The normal force acting on the block at position x is given by N = mg + F_y = mg + cx . The frictional force acting on the block is f_k = N = (mg + cx) . The magnitude of work done by the frictional force is the integral of this force from x=0 to x=d : |W| = ₀^ d f_k dx |W| = ₀^ d (mg + cx) dx |W| = [ mgx + cx^2 2 ]₀^ d |W| = ( mgd + cd^2 2 ) Answer: (mgd + cd^2 2 )