Question

A truck moves at 20 m/s carrying a 1000 kg box. Coefficient of friction between box and platform is \(\mu = 0.25\). Find the shortest stopping distance so that the box does not slip. (round off to nearest integer).

Hint:

To avoid slipping, deceleration must not exceed \(\mu g\). Use \(v^2=2as\) to find stopping distance.

Answer 1

Correct Answer: 79

Maximum deceleration without slipping:
\[ a_{\max} = \mu g = 0.25 \times 10 = 2.5\ \text{m/s}^2 \] Stopping distance using:
\[ v^2 = u^2 - 2 a s \] \[ 0 = 20^2 - 2(2.5)s \] \[ 400 = 5s \] \[ s = 80\ \text{m} \]