A 20.0-kg rock is sliding on a rough, horizontal surface at 8.00 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is 0.200. What average power is produced by friction as the rock stops?

Solution 52E Step 1: Consider the data given Mass of rock m = 20 kg Velocity of the down the inclined plane v = 8 m/s Kinetic friction M K = 0.200 Let us find the kinetic energy of the rock This is obtained by using KE = (1/2) mv 2 Substituting the values we get 2 KE = (1/2) × 20 kg × (8.0 m/s) KE = 640 J Thus we find kinetic energy as 640 J Step 2 : We shall find the force acting on the rock This is obtained as F = mg Substituting the values we get F = 20 kg × 9.81 F = 196.2 N Thus we have force on rock as 196.2 N Step 3: Let us find the kinetic friction of the rock K = M × F f K Substituting values we get K = 0.200 × 196.2 N f K = 39.24 N f Thus the kinetic friction of the rock is 39.24 N