A skier of mass m starts from rest and slides down a frictionless slope of lengt

Question

A skier of mass m starts from rest and slides down a frictionless slope of length d that is inclined at angle ? with the horizontal. Ignore air resistance. a) Calculate the work done by the gravity on the skier and the work done by the normal force on the skier? b) If the slope is not frictionless so that the skier has a final velocity v, calculate the work done by gravity, the work done by the normal force, the work done by friction, the force of friction, and the coefficient of kinetic friction?

Explanation / Answer

The slope is the hypotenuse of a right triangle with a 15.0° angle between the base and the hypotenuse. sin 15.0° = (Height of slope) ÷ (length of slope) sin 15.0° = (Height of slope) ÷ 27m Height of slope = 27 m * sin 15.0° = 6.988 m Work done by gravity = mass * gravity * height Work done by gravity = 66 kg* 9.8m/s^2 * 6.988 Work done by gravity = 4519.8 J = 4.52 kJ I found the work done by gravity on the skier and the work done by the normal force on the skier, which was 4.52 kJ and 0 kJ, respectively. Since the normal force is perpendicular to the direction of motion, no work is done by it. A force perpendicular to the direction of motion has no component in the direction of motion. Work done by normal force = 0

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