Question

A 5.0 kg 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 constant velocity for 10 cm. The work done by the applied horizontal force and the frictional force are respectively (take g=10m/s²)

Difficult

Solution

As velocity is constant work done by the applied force is only to overcome the frictional force.
so work done by the applied force is equal to $μ m g s = 0.5 \times 5 \times 10 \times 0.1 = 2.5 J b y f r i c t i o n a l f o r c e = - 2.5 J$

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