The work-energy theorem assumes that *F* and Δ*s* 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, Δ*s* 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 Δ*s*. The work done by the parallel component would then be . The work done by the perpendicular component would be zero since there is no displacement in that direction. In other words,

A few situations which are worth noting:

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

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

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

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

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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 **

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