Topic: Engineering Mechanics
A heavy block of mass m is slowly placed on a conveyer belt moving with speed v. If coefficient of friction between block and the belts is $\mu$, the block will slide on the belt through distance
- A. $\frac{v}{\mu g}$
- B. $\frac{v^{2}}{\sqrt{\mu g}}$
- C. $(\frac{v}{\mu g})^{2}$
- D. $\frac{v^{2}}{2\mu g}$
Answer: D. $\frac{v^{2}}{2\mu g}$
Explanation: Relative to the belt, the initial velocity of the block is $v$ and final velocity is $0$. The retardation due to the friction force is $a = \mu g$. Using the equation of motion $v^{2} - u^{2} = 2as$, we get $0 - v^{2} = 2(-\mu g)s$. Therefore, $s = \frac{v^{2}}{2\mu g}$.