A 58.0 kg skier is coasting down a 25º slope as the drawing shows. Near the top

Question

A 58.0 kg skier is coasting down a 25º slope as the drawing shows. Near the top of the slope, her speed is 3.6 m/s. A kinetic frictional force opposes her motion as she slides down the slope. Ignore air resistance.
a. Determine the skier’s acceleration as she moves down the slope.
b. How much work is done by the gravitational force if the skier moves a distance of 100.0 m down the slope?
c. If the coefficient of kinetic friction is 0.14, how much work is done by the frictional force as the skier moves a distance of 100.0 m down the slope?
d. What will the skiers speed be after moving 100.0 m down the slope?

Explanation / Answer

A) F=ma

mgsin-k.mgcos =m.a

a = g(sin-kcos)

B) W = mg(s.sin)

    W = 58.0(9.8)(100.sin 250)=2.40 x 104 J

C) Wf = -k.mgcos.s = -0.14.(58)(9.8)(cos25).100 = -7.21 x 103 J

D) mg.(s.sin)+ 0.5mv12 -k.mgcos.s = 0.5mv22

v2 = 24.34 m/s

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