A 79.0-kg skier starts from rest and slides down a 36.0-m frictionless slope that is inclined at an angle of 15.0° with the horizontal. Ignore air resistance.
(a) Calculate the work done by gravity on the skier and the work done by the normal force on the skier. Work done by gravity? Work done by normal force?
(b) If the slope is not frictionless so that the skier has a final velocity of 4 m/s, calculate the work done by gravity, the work done by the normal force, the work done by friction, the force of friction (assuming it is constant), and the coefficient of kinetic friction. Work done by gravity? Work done by normal force? Work done by friction?
a) Normal force is always perpendicular to the displacement so work done by it is zero
work done by gravity= mgsin(150)*36=7213.6 J
b) Work done by normal force is again zero due to the normal force
Work done by gravity is again the same as part a because it only depends on the force value, displacement and angle between them
But this time final K.E. energy is different and by work energy theorem, the difference between part b K.E. and a is the work done by friction
Work done by friction=.5*79*4*4 -7213.6=-6581.6 J
Friction force = Work/displacement = 6581.6/36~182.8 N
coefficient of kinetic friction = 182.8/79*9.8*cos(150)~0.244
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