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

\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.

\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.

\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.

Demonstration

Weight Lifters

Comics 

Atlas

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s