Conservation of Power and Energy

Question

A 13 kg block is slide from a height of 8.8 m down a friction-less incline. It comes to a flat surface and crosses a 7.8 m strip of sandpaper with a coefficient of friction of 0.01. After passing over the sandpaper, the block is traveling on a friction-less surface again. How fast is the block traveling? 

Answer 1

A 13 kg block slides from a height of 8.8 m 

energy at top of the height = Gravitational potential energy = mgh = 13*9.81*8.8 = 1122.26 J

energy at the bottom of the incline = energy at top of the height ( due to conservation of total mechanical energy)

Energy is not conserved at the bottom of the incline and the end of the sand paper, because some of the energy will spend to do work against the friction.

energy at the end of the sand paper = energy at the bottom of the incline - work done against the friction

work done against the friction = Friction force * Distance traveled

= coefficient of friction * normal force * Distance traveled

= 0.01* mg * Distance traveled

= 0.01 * 13 * 9.81 * 7.8

= 9.95 J

energy at the end of the sand paper = energy at the bottom of the incline - work done against the friction

= 1122.26 - 9.95 = 1112.31 J

this energy will transfer to the motion at the last surface

energy on the last surface = Kinetic energy = 1/2 m v²

1112.31 = 1/2 m v²

1112.31 = 1/2 *13* v²

v = √1112.31*2/13

v = 13.08 m/s