# Work done by friction over closed path

I am stuck thinking about work done by non-conservative forces. It is path dependent.

Let us consider an example.

A truck starts from rest and a block is kept on it. It accelerates for some time and then the driver applies brakes and the truck stops. In all this process, the block is moving with the truck. So what is the work done by the friction on the block?

Is it force time displacement ? Or is it zero as the change in kinetic energy over this time period is zero?

## 2 Answers

Short Answer: It is zero.

Long Answer:

Assume you are on a frame of reference of road (i.e you are standing on road) then the system is truck + block.

You observe both of them moving now looking closely friction is holding block on truck by directing itself in forward direction, using Newton's third law equal and opposite forces are acting on the truck. Thus work done by friction on whole system is:

$$W_f = W_{f \ on \ block} + W_{f \ on \ truck}$$ $$W_f = (F_{f \ on \ block} + F_{f \ on \ truck}) \cdot d$$ $$F_{f \ on \ block} = - F_{f \ on \ truck}$$ $$\therefore W_f = 0 \\ \mathbf{(on \ the \ system)}$$

Now, assume you are on truck block's displacement in your frame of reference block is stationary thus work done by friction is zero.

{Note: changing frame of reference is not easy in non-inertial frame}

Per the work energy theorem the net work done on an object equals its change in kinetic energy. Since the block begins and ends at rest its change in kinetic energy is zero. Therefore the net work done on the block by the static friction force is zero

When the truck accelerated the work done by friction was positive giving it kinetic energy since the force was in the same direction as the displacement of the block. During deceleration the work done by friction was negative taking away an equal amount of kinetic energy since the force was in the opposite direction as the displacement.

Hope this helps