Posted by **Lilly** on Tuesday, October 5, 2010 at 6:00pm.

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**, Tuesday, October 5, 2010 at 6:37pm
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**, Thursday, November 18, 2010 at 10:25am
3.108

- Physics -
**wafaa**, Saturday, April 30, 2011 at 12:33pm
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**, Sunday, November 11, 2012 at 11:12pm
the answer is shake your money maker

- Physics -
**Khai Lim**, Saturday, October 22, 2016 at 10:10pm
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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