A block of mass 2 kg is pushed up a rough inclined plane of angle 45^ at a constant velocity by a strictly horizontal force. The coefficient of kinetic friction between the block and the incline is 0.5 . The work done by the horizontal applied force as the block moves a distance of 5 m up the incline is: (Take g = 10 m/s ^2 )

Step-by-step solution

Let the horizontal applied force be F . We resolve the forces parallel and perpendicular to the inclined plane. The normal force N is given by the component of weight and the component of the horizontal force pushing into the incline: N = mg 45^ + F 45^ Since the block moves up the incline at a constant velocity, the net force along the incline is zero. The component of F up the incline balances the component of weight and the kinetic friction down the incline: F 45^ = mg 45^ + _k N F 45^ = mg 45^ + _k (mg 45^ + F