Understanding Work in Physics: Force and Displacement Relationship

[Vykan12]

I've always thought that the work done by a force is the scalar product of the force vector and the potential displacement vector it would have if no other forces were acting on it. My teacher says that it is really the product of the force vector and the resultant displacement regardless of what other forces act on the system, and that you'd get the same net work with either calculation.

I don't see how the way he defines it would make any sense. For example, if you derive the work done by gravity, you could find it to be just about anything depending on the external forces in question, rather than just mg(y_1 - y_2). So what is the true convention?

 

[tiny-tim]

Hi Vykan12!

Work done by a particular force is (integral of) that force times the actual displacement (of the point of application of the force, if it's not a point body).

What the displacement would be if there were no other forces is irrelevant.

 

[Doc Al]

For example, if you derive the work done by gravity, you could find it to be just about anything depending on the external forces in question, rather than just mg(y_1 - y_2).

No, regardless of the path the object takes the work done by gravity will always be mg(y_1 - y_2).

 

[Vykan12]

I can see that's where the properties of conservative forces come into play. Interesting.