Question

A child starts from rest and slides down a snow-covered hill with a slope angle of 47° from a height of 1.8m above the bottom of the hill. The speed of the child at the bottom of the hill is 3.1m/s. Find the coefficient of kinetic friction between the hill and the child.

My answer is 0.54, is that correct?

Answer 1

Potential energy at top = m g h = 1.8 m g
Kinetic energy at bottom = (1/2) m v^2 = .5 m(3.1)^2
Energy loss = L = m(1.8 g -.5*3.1^2)
work done by friction = L = F * d = F*1.8/sin 47
normal force = m g cos 47
F = mu * normal force = mu m g cos 47
so
work done by friction = mu m g cos 47 * 1.8 /sin 47
so in the end
mu g cos 47 * 1.8 /sin 47 = (1.8 g -.5*3.1^2)

Answer 2

I get closer to .78 but do not have calculator handy

Answer 3

Yes, that's what I'm getting too, thank you!