# Work in physics

1. Jun 16, 2013

Hello, if an object is moved to the right and then to the left back to its original spot, is work 0. Or would work still have a value?

Work = Force * Delta x.

It just doesn't seem to make sense to say that there is no work done even when you moved it around.

Last edited: Jun 16, 2013
2. Jun 16, 2013

### QED Andrew

In physics, work has a much more precise definition than the everyday meaning of work. You seem to be confusing the two definitions. As you can see from the equation in your post, if either the magnitude of the component of the net force parallel to the displacement of the object or the displacement of the object is zero, then the net work done on the object is zero. This result is nothing more than a direct consequence of the definition of work in physics.

3. Jun 16, 2013

### Staff: Mentor

The equation for work is generally specified as a dot product: $W = \vec{F} \cdot \vec{Δx}$

If the force that moves your object to the right is also directed to the right, and if the force that moves it back again is directed leftwards in the direction of the motion that returns it to its origin, then in both cases positive work will be done on the object and the total work will be a positive, nonzero value.

4. Jun 17, 2013

### CWatters

If you haven't covered dot products yet... What happens if you look at it in two parts....

Work = (work required to move it right) + (work required to move it left)
Work = (+force * +deltaX) + (-force * -deltaX)
= 2 * force * deltaX

The "-" signs are there because the direction of the force and displacement has changed.

5. Jun 17, 2013

### QED Andrew

Is it fair to assume so much about the force(s) acting on the object? It isn't difficult to imagine a case in which variable forces produce the required motion but the net work done on the object is zero.

6. Jun 17, 2013

### Staff: Mentor

Sure, take circular motion for example. The centripetal force acts always towards the center of the motion and is always at right angles to the direction of motion. Thus there is no component of the force acting along the direction of motion, hence the work done is zero yet the object travels away from and returns to a given location repeatedly.

When more complicated variable forces are involved, the usual approach is to integrate $\vec{F}\cdot\vec{ds}$ over the path the object takes.

7. Jun 18, 2013

### CWatters

No clearly it's not. I only offered that as a simple explanation of the particular example described by the OP.

Is there a way to explain the general case without reference to dot products?