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A 62 kg skier slides down a 25 degree slope, and the kinetic friction on the skier is measured to be 45.0 N. Using g = 9.80 m/s squared, what is the coefficient of kinetic friction?
A0.045
B0.204
C0.164
D0.082
Answer & Solution
Correct answer: D. 0.082
1. The normal force on the slope equals the component of weight perpendicular to the slope, N = mg cos(25 degrees).
2. Substituting values, N = (62 kg)(9.80 m/s squared)(cos 25 degrees) = 607.6 N times 0.9063, which is about 550.8 N.
3. Kinetic friction is f_k = mu_k N, so mu_k = f_k / N = 45.0 N / 550.8 N.
4. That division gives mu_k of about 0.082.
5. 0.164 doubles the correct value, a trap from forgetting the cosine factor and using the full weight of 607.6 N with a different rounding error.
6. 0.204 would come from dividing 45.0 N by the sine component of weight rather than the cosine, mixing up which component is normal to the slope.
_Source: OpenStax College Physics (CC BY 4.0), Ch 5 "Further Applications of Newton's Laws: Friction, Drag, and Elasticity", section 5.1 Friction_
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