Question

A box of mass 20kg is pushed along a rough floor with a velocity 2m/s and then let go

Answer 1

The frictional force acting on the box is 4 N

Explanation:

Mass of the box M = 20 kg

Initial Velocity u = 2 m/s

Final velocity v = 0

Distance travelled s = 5 m

USing the third equation of motion

v^2=u^2+2asv

2

=u

2

+2as

0^2=2^2+2\times a\times 50

2

=2

2

+2×a×5

\implies a=-\frac{4}{2\times 5}⟹a=−

2×5

4

\implies a=-0.4⟹a=−0.4 m/s^2

This retardation will be due to force of friction

Therefore, the frictional force acting

F=MaF=Ma

F=10\times 0.4F=10×0.4 N

\implies F=4⟹F=4 N

Hope this helps.