anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Also, this is dealing with vectors F and magnitude
anonymous
  • anonymous
Ok, This is what I got. One sec
mathmate
  • mathmate
\(work~done, W=\textbf{F.D}=FDcos(\theta)\) where \(\theta\) is the angle between \(\textbf{F}~and~\textbf{D}\). W is a scalar.

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More answers

mathmate
  • mathmate
|dw:1435713093544:dw|
anonymous
  • anonymous
F = 5(cos ai + sin aj) cos 45 + sin 45 \[ \huge 5(\frac{\sqrt{2}}{2}i + \frac{\sqrt{2}}{2}j ) \] \[ \huge \frac{5\sqrt{2}}{2}i + \frac{5\sqrt{2}}{2}j \]
anonymous
  • anonymous
Now I guess the W is the magnitude of force to the distance?
anonymous
  • anonymous
Or magnitude time distance
mathmate
  • mathmate
It is a dot product. So you need to put the distance as a vector as well, then proceed with the dot product!
anonymous
  • 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?
dan815
  • dan815
yep that looks good
anonymous
  • anonymous
Thanks!
mathmate
  • mathmate
Yep, that's right.

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