A ski starts from rest and slides down a 27° incline 110 m long. Use energy methods.?
(a) If the coefficient of friction is 0.09, what is the ski's speed at the base of the incline?
(b) If the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level?
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Verified answer
a/
initial PE - friction work = final KE
mgh - µmgdcosΘ = ½mv²
The ski's mass cancels --
9.8m/s² * 110m*sin27º - 0.09 * 9.8m/s² * 110m * cos27º = v²/2
solves to
v = 28 m/s
to two significant digits
b/
KE becomes friction work:
½mv² = µmgx
x = v² / (2µg) = (28m/s)² / (2 * 0.09 * 9.8m/s²) = 457 m ≈ 460 m
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