Question

An 80.0 kg rollerblader is at rest at the top of a 100 m hill at a 7o incline. The coefficient of friction on the hill is 0.110. What is the roller blader's kinetic energy at the bottom of the hill?

Answer 1

I assume 100 meters is the height of the hill, not the distance down the slope.

loss of potential energy
= 80 * 9.81 * 100

normal force on slope = 80 * 9.81 *cos 70

friction force = 0.110 * normal force
= 0.110 * 80 * 9,81 * cos 70

distance moved down slope
= 100 /sin 70

so work done against friction =
0.110*80*9,81*cos 70*100/sin 70

ke = m g h - work done on friction

(1/2) 80 v^2 = 80 * 9.81 * 100 - 0.110*80*9,81*cos 70*100/sin 70

cancel 80. Solve for v