A block A with mass $m$ is pushed across a horizontal tabletop with a constant velocity by a horizontal force of magnitude $F$. What is the coefficient of kinetic friction between block A and the tabletop?

• A. $\frac{m}{Fg}$
• B. $\frac{F}{mg}$
• C. $\frac{g}{Fm}$
• D. $\frac{F}{g}$

Answer: (B)

Verified step by step guidance

1. Identify the forces acting on the block: the applied horizontal force \(F\), the kinetic friction force \(f_k\), the gravitational force \(mg\) downward, and the normal force \(N\) upward.
2. Since the block moves with constant velocity, the net force in the horizontal direction is zero. Therefore, the applied force \(F\) balances the kinetic friction force \(f_k\), so \(F = f_k\).
3. The kinetic friction force is related to the normal force by \(f_k = \mu_k N\), where \(\mu_k\) is the coefficient of kinetic friction.
4. On a horizontal surface, the normal force \(N\) equals the weight of the block, so \(N = mg\).
5. Combine these relations to solve for the coefficient of kinetic friction: \(\mu_k = \frac{f_k}{N} = \frac{F}{mg}\).