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The same 100 kg crate from the previous problem is now sliding, where the coefficient of kinetic friction is 0.30. Using g = 9.80 m/s squared, what force keeps it sliding at a constant speed?
A441 N
B196 N
C294 N
D980 N
Answer & Solution
Correct answer: C. 294 N
1. The normal force is unchanged from the static case, N = mg = (100 kg)(9.80 m/s squared) = 980 N.
2. Once sliding, kinetic friction is f_k = mu_k N, an equation rather than a maximum.
3. Substituting mu_k = 0.30 gives f_k = (0.30)(980 N) = 294 N.
4. A constant speed means the net force is zero, so the applied force must exactly cancel this kinetic friction, giving 294 N.
5. 441 N reuses the static coefficient of 0.45 by mistake, and 980 N is again the bare normal force with no coefficient applied.
6. 196 N would come from mu_k = 0.20, a coefficient that does not appear in this problem at all.
_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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