Work: Vector Integration

**Aryth** · September 18th 2008, 07:18 PM

The force function $\bold{F} = -\bold{\hat{x}}kx - \bold{\hat{y}}ky$ is using Cartesian coordinates.

Find the work done from (1,1) to (4,4) using the following path:

$(1,1) \to (1,4) \to (4,4)$

So, what I did was:

$\int^{4,4}_{1,1} \bold{F}\cdot ~d\bold{r}$

$= \int^{4,1}_{1,1} \bold{F}\cdot ~d\bold{r} + \int^{4,4}_{4,1} \bold{F}\cdot ~d\bold{r}$

$= \int^{4,1}_{1,1} (-kx ~dx - ky ~dy) + \int^{4,4}_{4,1} (-kx ~dx - ky ~dy)$

$= \int^4_1 (-kx ~dx) + \int^1_1 (-ky ~dy) + \int^4_4 (-kx ~dx) + \int^4_1 (-ky ~dy)$

$= -k\left(\int^4_1 x ~dx + \int^1_1 y ~dy + \int^4_4 x ~dx + \int^4_1 y ~dy\right)$

$= -k\left[\left(8 - \frac{1}{2}\right) + \left(8 - \frac{1}{2}\right)\right]$

$= -k(15) = -15k$

Is this right?

**icemanfan** · September 18th 2008, 07:34 PM

It seems to me that if you are heading in the x-direction the work done will only depend on the x component of the force, and if you are heading in the y direction the work done will only depend on the y-component of the force, so you can add $\int _1 ^4 f(y) dy + \int _1 ^4 f(x) dx$ where f(y) is the y-component of force and f(x) is the x-component of force. Your answer looks right to me.

**Jhevon** · September 18th 2008, 07:54 PM

i agree. your answer is correct. i've never seen that method before, though. it seems to make sense

**Aryth** · September 18th 2008, 08:07 PM

I've never seen this method either... But it's the way me and my friends interpreted it given this example:

The force exerted on a body is $\bold{F} = -\bold{\hat{x}}y + \bold{\hat{y}}x$ . The problem is to calculate the work done going from the origin to the point (1,1):

$W = \int^{1,1}_{0,0} \bold{F} \cdot ~d\bold{r} = \int^{1,1}_{0,0} (-y ~dx + x ~dy)$

Separating the integrals, we obtain:

$W = -\int^1_0 y ~dx + \int^1_0 x ~dy$

The first integral cannot be evaluated until we specify the values of x as y ranges from 0 to 1. Likewise, the second integral requires x as a function of y. Consider the first path:

$(0,0) \to (1,0) \to (1,1)$

Then:

$W = -\int^1_0 0 ~dx + \int^1_0 1 ~dy = 1$

since y = 0 along the first segment of the path and x = 1 along the second. If we select the path:

$0,0 \to (0,1) \to (1,1)$

then the integral gives $W = -1$ . For this force the work done depends on the choice of path.

That's pretty much all we had to go by. No professor, no class, just us and a book...

We did have this relation:

$W = \int \bold{F}\cdot ~d\bold{r} = \int F_x(x,y,z) ~dx + \int F_y(x,y,z) ~dy + \int F_z(x,y,z) ~dz$

Using that, the separation of the integrals follows.

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