A child insists on going sledding on a barely snow-covered hill. The child starts
from rest at the top of the 60 m long hill which is inclined at an angle of 30o to the horizontal, and arrives at the bottom 8.0 s later. What is the coefficient of kinetic friction between the hill and the sled?
Assume that the acceleration is constant, such that
60 m = (a/2)t^2
a = 120/(8)^2 = 1.875 m/s^2
Then use Newton's second law in the form
Fnet = m*g sin30 - m*g*cos30*u = m*a
Note that the mass cancels out, which is good since they did not provide a value.
With the value of a that you now know, the friction coefficient u can be calculated.
0.50 g - 0.866 u*g = 1.875 m/s^2
8.49 u = 4.90 -1.875 = 3.02 m/s^2
u = 0.356posted by drwls