Asked 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?
Answered by bobpursley
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
Answered by Anonymous
3.108
Answered by wafaa
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?
Answered by george
the answer is shake your money maker
Answered by Khai Lim
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.
Answered by Sayam Sriram
Very well-versed question, thank you for your contribution to the physics world!

-A concerned citizen
Answered by Sayan Sriram
Sorry on the typo above!
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