## anonymous one year ago Find the work W done by a force of 5 pound acting in the direction of 45 degree to the horizontal in moving an object 7 feet from (0,0) to (7,0) W = ? foot-pounds

1. anonymous

Also, this is dealing with vectors F and magnitude

2. anonymous

Ok, This is what I got. One sec

3. mathmate

$$work~done, W=\textbf{F.D}=FDcos(\theta)$$ where $$\theta$$ is the angle between $$\textbf{F}~and~\textbf{D}$$. W is a scalar.

4. mathmate

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5. anonymous

F = 5(cos ai + sin aj) cos 45 + sin 45 $\huge 5(\frac{\sqrt{2}}{2}i + \frac{\sqrt{2}}{2}j )$ $\huge \frac{5\sqrt{2}}{2}i + \frac{5\sqrt{2}}{2}j$

6. anonymous

Now I guess the W is the magnitude of force to the distance?

7. anonymous

Or magnitude time distance

8. mathmate

It is a dot product. So you need to put the distance as a vector as well, then proceed with the dot product!

9. anonymous

So..... $\huge (\frac{5\sqrt{2}}{2}i + \frac{5\sqrt{2}}{2}j) * 7i$ $\huge (\frac{5\sqrt{2}}{2}(7) + \frac{5\sqrt{2}}{2}(0))$ $\huge (\frac{5\sqrt{2}}{2}(7) + \frac{5\sqrt{2}}{2}(0))$ $\huge \frac{5\sqrt{2}}{2}(7)$ $\huge W = \frac{35\sqrt{2}}{2}$ Correct?

10. dan815

yep that looks good

11. anonymous

Thanks!

12. mathmate

Yep, that's right.