Athenian* Christina Curtis writes in with a frequently asked question about calculating work.
She's from Athens, Georgia, not the one with the
Consider a box of mass m being pushed up a frictionless incline at constant speed by a person. We know the angle of the incline θ, the distance along the incline d, and the vertical height of the incline h. What force and what distance do we use to calculate the work done by the person?
It is valid here to use the definition of work as force times distance, because all of the forces involved are unchanging. If the forces were varying, as on a curved track or for something in simple harmonic motion, you'd have to use the area under a force vs. distance graph, or go straight to the work-energy theorem.
When using work = force x distance, the distance you use must be parallel to the force involved. (Or, equivalently, the component of force you use must be parallel to the direction of displacement.)
One approach, then, is to use the vertical displacement multiplied by the vertical component of the force applied by the person. We know the vertical displacement is h, and the vertical component of the person's force is simply mg. The work done by the person is mgh.
A second approach is to use the distance along the incline multiplied by the force of the person on the block. To get equilibrium in the direction parallel to the incline, the person pushes with a force mgsinθ. So the work done by the person is (mgsinθ)(d).
Wait a second! How do we get two different answers using two different approaches?
We don't. In this geometry, the sine of the angle θ is h / d. Using that relationship, you can see that the two expressions we've derived above for work are equivalent. Either approach -- using the force component that's parallel to displacement, or using the displacement component that's parallel to force -- works.
But it might be easiest to use the work-energy theorem.
Students tend to default to work = force x distance. But the work-energy theorem is more generally valid, and is often less confusing.
The work-energy theorem says that work done by an external force is equal to the change in kinetic energy plus change in potential energy plus the change in internal energy. Here, consider the block-earth system. The block has no change in kinetic energy 'cause it moves at constant speed. It has no change in internal energy 'cause it doesn't change its temperature, and doesn't have anything inside it storing energy through rotation or something. So the work done by an external force -- that is, the work done by the person -- is just equal to the change in the block's potential energy. That's mgh.
Ideally, we'd get our students to default to using the work-energy theorem to calculate work done, then only use force x distance if they have to.