Physics
posted by Lilly .
A skier traveling 12.0 m/s reaches the foot of a steady upward 18° incline and glides 18.2 m up along this slope before coming to rest. What was the average coefficient of friction?

initial KE= work done on gravity+workdone on friction.
1/2 m 12^2=mgh/sin18 + mu*mg*sin18*h/sin18
72=gh/sin18 + mu*g*h
we know h=18.2/sin18
so, solve for mu 
3.108

a skier traveling 12.0 m/s reaches the foot of a steady up ward 18.0 incline and glides 12.2m up along this slop before coming to rest. what the average cofficient of friction?

the answer is shake your money maker

KE = PE + W_friction
1/2 * mv^2 = mgh + mu*mgcos(theta)d
 m is cancel out
 sin(theta) = heigh/ distance( inclide) ===> h = d*sin(theta)
1/2 v^2 = g (d*sin(theta)) + mu* (g*cos(theta))*d
now you can solve for mu ^^ Good luck.
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A skier traveling 12.0 m/s reaches the foot of a steady upward 18 degree incline and glides 12.2 m up along this slope before coming to rest. What was the average coefficient of friction?