Question

A box of mass 20 kg is pushed along a rough floor with a velocity 2 $\frac{m}{s}$ and then let go. The box moves 5 m 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 = 20 kg; Initial velocity u = 2 $\frac{m}{s}$; Final velocity v = 0 $\frac{m}{s}$;
Distance travelled s = 5m; Frictional force f = ?
As we know, $v^2-u^2=2as$
$\therefore a=\frac{v^2-u^2}{25}=\frac{-4}{10}=0.4\frac{m}{s}$
Therefore, frictional force f= m $\times$ a = 20 $\times$ 0.4
= 8 N