Question

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

Answer 1

Also, this is dealing with vectors F and magnitude

Answer 2

Ok, This is what I got. One sec

Answer 3

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

Answer 4

F = 5(cos ai + sin aj)

cos 45 + sin 45

\[ \huge 5(\frac{\sqrt{2}}{2}i + \frac{\sqrt{2}}{2}j ) \]
\[ \huge \frac{5\sqrt{2}}{2}i + \frac{5\sqrt{2}}{2}j \]

Answer 5

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

Answer 6

Or magnitude time distance

Answer 7

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

Answer 8

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?

Answer 9

yep that looks good

Answer 10

Thanks!

Answer 11

Yep, that's right.