A semi-circle is given by the function [tex]y=R cos(\frac{x\pi}{2R})[/tex]. An attractive force F at [tex]x = R[/tex] and [tex]y = 0[/tex] such that [tex]F(x, y) = F(R, 0) = 0 [/tex] and [tex]F(x, y) = F(-R, 0) = F_{0} [/tex]. This force along the semi-circle, relative to its source, is given by [tex]F(\theta) = F_{0} sin(\theta/2)[/tex]. What amount of work is done against the force when a particle moves along the circle from [tex]\theta = 0[/tex] to [tex]\theta = \pi[/tex].(adsbygoogle = window.adsbygoogle || []).push({});

To find the work I have the equation

[tex]W = - \int^{\pi}_{0} F dr[/tex]

For the force I have the equation

[tex]F= F_{0} sin(\theta/2)[/tex]

For the distance from the particle to the orgin of the force I have

[tex]2 R sin(\theta/2)[/tex]

so for dr I have

[tex]2 R sin(\theta/2) d\theta[/tex]

Putting this all tegether I get

[tex]W = - \int^{\pi}_{0} F_{0} sin(\theta/2) 2 R sin(\theta/2) d\theta[/tex]

This rearanges to

[tex]W = -F_{0} R \int^{\pi}_{0} \left[1-cos(\theta) \right] d\theta[/tex]

Integrating I get

[tex]W = -F_{0} R \left[ \theta-sin(\theta) \right]^{\pi}_{0}[/tex]

This works out to

[tex]W = -F_{0} R \pi[/tex]

The answer I am suppose to get is

[tex]W = -F_{0} R[/tex]

What did I do wrong?

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# Problems with work and F*dr

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