Question

A wooden crate is pushed across a concrete floor at $5m/s$ and released. It slides to a stop after moving a short distance. The same crate is filled until it weighs twice as much as it did previously and again slid across the floor at $5m/s$ and released. The stopping distance for the crate will be

• A. $1/4$ as far
• B. $1/2$ as far
• C. The same distance
• D. Twice as far
• E. Four times as far

Answer 1

Answer: (C)

The same distance

The deceleration of the crate is $-\mu g$ which is independent of the mass of the crate.

The stopping distance can be found by $0^2-v^2=2(-\mu g)s$

$⟹s=\frac{v^2}{2\mu g}$ independent of the mass of the crate.

Hence even when the mass of the crate is doubled, the stopping distance remains the same.

Correct answer is option C.