Question

A toboggan, with rider, has a combined mass of 60kg at the top of a 5.0m hill. They are given an initial push of 2.0m/s. The distance down the hill was 20m. The velocity at the bottom of the hill is 10m/s. If the actual speed at the bottom was 9m/s, what would have been the average coefficient of friction between the toboggan and snow?

Answer 1

initial energy+PE added=final KE+frictionwork

1/2 *60*2^2+60*9.8*5=1/2 60*9^2+60*9.8*mu*cosineTheta
solve for mu, the coefficent of friction.
You actually have to know the slope of the hill.
tan theta=5/20=1/4
theta=arctan(.25)=14 degrees
cosTheta(14deg)= 0.97

Answer 2

I suspect you are assuming that g = 10 m/s^2
total energy at top = (1/2) m (2)^2 + m(2)(5) = 2 m + 50 m = 52 m Joules

energy at bottom at 10 m/s = (1/2) m (100) = 50 m Joules

If the speed at the bottom was only 9m/s
then
energy at bottom = (1/2) m (81) = 40.5 m Joules
52 - 40.5 = 11.5 m Joules lost to friction

Friction force * distance traveled = work done against friction
F * 20 = 11.5
F = .575 Newtons
that is
mu m g cos(slope angle) = .575
sin slope = 5/20
so
cos slope = .968
mu (60)(10)(.968) = .575
mu = 9.9 * 10^-4 = 0.00099

Answer 3

Friction force * distance traveled = work done against friction
F * 20 = 11.5 m
F = .575 m Newtons
that is
mu m g cos(slope angle) = .575 m
sin slope = 5/20
so
cos slope = .968
mu (60)(10)(.968) = .575 (60)
mu = .0594
(my previous answer times m, 60 kg

Answer 4

In other words the answer does not depend on the mass at all.