The 10-kg block slides down 2 m on the rough surface with kinetic friction coefficient μk = 0.2. What is the work done by the friction force?

153.2 J

Explanation:

Let's first list our given parameters;

mass (m) of the block = 10 kg

which slides down ( i.e displacement) = 2 m

kinetic coefficient of friction (μk) = 0.2

In the diagram shown below;  if we take an integral look at the component of force in the direction of the displacement; we have

Fcos 40°

100 (cos 40°)

76.60 N

Workdone by the friction force can now be determined as:

W = × displacement

W = 76.60 × 2

W = 153.2 J

∴  the work done by the friction force = 153.2 J

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