Question

A $5.0kg$ box rests on a horizontal surface. The coefficient of kinetic friction between the box and the surface is $0.5$. A horizontal force pulls the box at a constant velocity for $10cm$. The work done by the applied horizontal force and the frictional force are respectively (take $g=10m/s^2$):

• A. 2.5 J and 2.5 J
• B. zero and 2.5 J
• C. 2.5 J and zero
• D. 2.5 J and - 2.5J

Answer 1

Answer: (D)

2.5 J and - 2.5J

Given, Mass of the box $=5Kg,$

Coefficient of kinetic friction$\mu =0.5$

Displacement $=10cm$

Since box moves with constant velocity means acceleration = 0

Force applied$F=$ Frictional force $=\mu \times mg$ = 25 N

work done by force$=F\times S=25\times 0.1=2.5J$

Applying work-energy theorem

$W_{friction}+W_{force}=\Delta KE$

Since no change in velocity so $\Delta KE=0$

hence $W_{friction}=-W_{force}=-2.5J$

hence option (d).