Calculate the work done by a force

[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??

 

[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.

 

[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)?

 

[Dale]

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

 

[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!

 

[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.

 

[don_anon25]

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

 

[FredGarvin]

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