Question

A body is coming with a velocity of $72 \,km/h$ on a rough horizontal surface of coefficient of friction $0.5.$ If the acceleration due to gravity is $ 10\,m/s^{2}, $ find the minimum distance it can be stopped.

• A. 400 m
• B. 40 m
• C. 0.40 m
• D. 4 m

Answer 1

Answer: (B)

Given, $u=72 \,km / h =20 \,m / s$
$a=\mu g=0.5 \times 10 \,m / s ^{2}$
From $v^{2}=u^{2}-2 \,as$
$\therefore (0)^{2}=(20)^{2}-2 \times 0.5 \times 10 \times s$
$\therefore s=\frac{20 \times 20}{2 \times 0.5 \times 10}$
or $s=40 \,m$