A block of mass m is sliding down an inclined plane with constant speed. At a certain instant t 0 , its height above the ground is h . The coefficient of kinetic friction between the block and the the plane is μ . If the block reaches the ground at a later instant t g , then the energy dissipated by friction in the time interval t g - t 0 is,

Step-by-step solution

Let u be the constant speed of the block. According to the work-energy theorem, the change in kinetic energy will be equal to the work done by the forces acting on the block in sliding down. W net = ∆ K . E Here, ∆ K . E = 1 2 m u 2 = 0 as the speed is constant. Thus, W f + W mg = 0 , where W f and W mg are the work done by frictional and gravitational forces. W f = - W mg = - m g h , here g is the acceleration due to gravity and h is height. Hence, energy dissipated by friction is m g h .