A question in the derivation of work done

1. Mar 22, 2013

ehabmozart

In our formulation of work done by a Force we say it is ∫ F.dl ... My knowledge of integration means adding Infinitesimal parts. We write dl because it is the smallest part of the distance moved. My question is why have we written F and not dF... Thanks for clarifying!

2. Mar 22, 2013

Staff: Mentor

The force doesn't get infinitesimally small. It remains finite as it acts over an infinitesimal distance.

3. Mar 22, 2013

ehabmozart

But not the full force is applied to this small part

4. Mar 22, 2013

Staff: Mentor

Sure it is.

Say I exert a force of 100 N over a distance of 1 meter. I'm exerting the full 100 N over every bit of that distance.

5. Mar 23, 2013

ehabmozart

What if the force is varying??

6. Mar 23, 2013

Staff: Mentor

That's when you integrate. If the force is constant (and the object moves in a straight line), you simply have $W = \vec F \cdot \vec {\Delta l}$.

7. Mar 23, 2013

Staff: Mentor

As jtbell said, that's why you integrate.

But the force at any point is whatever it is, not some infinitesimal.

Say I push with some varying force for a distance of 1 meter. You can break that path up into as many small sections as you like. (When you integrate, you are making them infinitesimally small.) At any point in the path, the force has some value F(x) as it pushes through the tiny distance dx.

By analogy, think of calculating the distance traveled by something moving with some speed v:
X = V*t
Expressing it as a differential, you get dX = V*dt, not dX = dV*dt. Over the tiniest interval of time, the speed does not go to zero.

Last edited: Mar 23, 2013