Conservation of Energy - Physics?

Question

A factory worker accidentally releases a 210 kg crate that was being held at rest at the top of a ramp that is 4.1 m long and inclined at 30° to the horizontal. The coefficient of kinetic friction between the crate and the ramp, and between the crate and the horizontal factory floor, is 0.23. (a) How fast is the crate moving as it reaches the bottom of the ramp? (b) How far will it subsequently slide across the floor? (Assume that the crate's kinetic energy does not change as it moves from the ramp onto the floor.)

Answer 1

Potential energy Pe at the top of the ramp gets converted to kinetic energy Ke and work against friction Wf

Pe=Ke+Wf

Ke=0.5 m V^2
Pe=mgh
Wf= fs= u mg cos(A) s
u - coefficient of friction
mg cos(A) is the normal to the surface component of weight
s- distance covered

Now
mgh=0.5 m V^2 + u mg cos(A) s

V=sqrt(2g( h - ucos(A) s))

b) Ke=Wf
0.5 m V^2 = u mg s (note cos(A) now = 1 since angle A=0)

s= 0.5 V^2/ (ug)

Funny how things work out. We didn't even used the mass of the crate.