# Calculate the work done by a force

1. Dec 13, 2005

### don_anon25

I am asked to calculate the work done by a force as it moves around a path. The force is F = b(1-x^2/a^2)j. The path is a rectangle with coordinates at (0,0); (0,L); (a,L); (a,0). The force moves clockwise around the path beginning at the origin. A diagram is attached.

I know work is the integral of F dot dr.
So for the first path I should have the the force F=b(1-x^2/a^2)j dotted with Lj (the path from the origin to point (0,L)). The integral is thus bL (1 - x^2/a^2) dy with limits from y=0 to y=L. Is this the right approach? If not, can someone please point me in the right direction??

#### Attached Files:

• ###### work.bmp
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2. Dec 13, 2005

### Physics Monkey

You have an extra factor of L that you don't need (look at the units). The work is the integral of $$\vec{F}\cdot d\vec{r}$$ along the path. For example, the first segment has $$d\vec{r} = dy \hat{j}$$. You have to figure out what $$d\vec{r}$$ is for each of the four segments.

3. Dec 14, 2005

### don_anon25

So, for the first segment, dr = dyj. For the second segment, dr = dx i.
For the third segment, dr = -dyj. For the fourth segment, dr = -dx i. Is this correct? Are the limits on my integration correct as well?

Also, should the answer be 0 (closed path, conservative force...not sure if the force is conservative though)?

4. Dec 14, 2005

### Staff: Mentor

If the answer is zero then the force is conservative, but not all forces are conservative so you can't use that as a check here. (Your dr vectors are correct).

-Dale

5. Dec 15, 2005

### don_anon25

I get an answer of 2bL (1- x^2/a^2). This does not seem correct to me, since it contains an x^2 term? Is this right? Is there a substitution I can make for x? x=a or x=L, for instance? This problem is driving me crazy...any help greatly appreciated!

6. Dec 15, 2005

### FredGarvin

You can eliminate two of the legs from your problem since the force is in the $$\hat{j}$$ direction.

In segment 1, x=0. In segment 3, x=a.

Last edited: Dec 15, 2005
7. Dec 15, 2005

### don_anon25

Yes...I have the forces for the dx direction to be zero. I'm still doing something wrong though?

8. Dec 15, 2005

### FredGarvin

Did you substitute in the values for x that I just edited into my last post?