5.1.2 Work Done by a Force

The work-energy theorem $W=F\Delta s$ assumes that F and Δs are in the same direction. What if they are not?

For example, in the scenario below, a box is being pulled along by a force F. Because other forces are involved (weight, normal contact, and possibly friction) here, the box does not move in the same direction as F. Instead, Δs of the box and F are misaligned by the angle θ.

In such scenarios, we can always resolve F into two components: parallel and perpendicular to Δs. The work done by the parallel component $F\cos \theta$ would then be $F\Delta s\cos \theta$ . The work done by the perpendicular component $F\sin \theta$ would be zero since there is no displacement in that direction. In other words,

$W=F\Delta s\cos \theta$

A few situations which are worth noting:

$\theta =0{}^\circ$

If the mass moves in the direction of the force, we have the original work-energy theorem!

$W=F\Delta s$

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$\theta =180{}^\circ$

If the mass moves in the opposite direction as the force, the force is doing negative work to the mass!

$W=-F\Delta s$

This is because the force is a retardation force. The negative work done by F represents the decrease in energy caused by the force.

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$\theta =90{}^\circ$

If the mass moves in a direction perpendicular to the force, the force is doing zero work to the mass!

$W=0$

This scenario must always involve other forces besides F. Any changes in the KE of the mass must be due to other forces, not F.

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$\Delta s=0$

If the mass does not move, the force is doing zero work to the mass!

$W=0$ .

Again, this scenario must involve other forces besides F. All these forces acting together on the mass, end up producing zero acceleration to the mass, and therefore does zero work.

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Demonstration

Weight Lifters

Comics

Atlas