Derive the relation for work done by a force inclined with the displacement?
Difficulty: Medium
Let the force F is making an angle q with the surface on the body is moved.
Resolving F into its perpendicular components Fx and Fy as;
$F_{x}=F\cos\theta$
$F_{y}=F\sin\theta$
In case when force and displacement are not parallel then only the x-component $F_{x}$ parallel to the surface causes the body to move on the surface and not the y - component $F_{y}$
Hence W = $F_{x}\times S$
$W= \left(F\cos\theta\right)S$
$W = FS \cos\theta$
Mini Exercise
A crate is moved by pulling the rope attached to it. It moves 10 m on a straight horizontal road by a force of 100 N. How much work will be done if?
1. The rope is parallel to the road
Solution: Force = 100 N
Distance = S = 10 m
Work = W = ?
$W=F\times S$
$W = 100 x 10 = 1000 J$
2. The rope is making an angle of 300 with the road.
Solution: Force = 100 N
Distance = S = 10 m
$\theta =30\circ$
Work = W = ?
$W=FS cos\theta$
$W = 100\times10 cos(30\circ)$
$W = 1000\times (0.866) = 866 J$
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