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A force of 10lb is used to move a box across a horizontal floor distance of 5 ft. If the force makes an angle of 30 degrees with the floor,how much work is done?
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@CShrix
\[\huge \text{W}=Fd \cos(\theta)\]
The cosine is involved because we need to find the component that is parallel to the force being applied.
so we need to convert LB TO KG?
and the ft too?
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Hmm, I would probably convert it.
ok, lets move on then...
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