Posted by Lilly on .
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?

Physics 
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 
Physics 
Anonymous,
3.108

Physics 
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?

Physics 
george,
the answer is shake your money maker

Physics 
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.