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 belt is mu, the block will slide on the belt through distance
- A. v / (mu * g)
- B. v² / $\sqrt{mu * g}$
- C. (v / (mu * g))^2
- D. v² / (2 * mu * g)
Answer: D. v² / (2 * mu * g)
Explanation: Mass of block = m, speed = v. Retardation due to friction = -mu * g. Using v² - u² = 2as, we have v² - 0 = 2 * mu * g * s, which gives s = v² / (2 * mu * g).