Question

A box of mass $20kg$ is moving with a velocity of $2m{s}^{-1}$ on a rough floor. The box moves $5m$ on the floor before coming to rest. What must be the frictional force acting on the box?

• A. 4 N
• B. 2 N
• C. 20 N
• D. 8 N

Answer 1

The correct option is D 8 N

Given: mass of box , $m=20kg$, initial velocity, $u=2m{s}^{-1}$, final velocity, $v=0m{s}^{-1}$ and distance travelled, $s=5m$.
Let frictional force be $F$.
From the third equation of motion, $v^2-u^2=2as$, we get $a=\frac{v^2-u^2}{2s}=\frac{0-2^2}{2\times 5}=0.4m{s}^{-2}$
$\therefore$ Frictional force, $F=m\times a=20\times 0.4=8N$