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A force F=(10+0.50x) acts on a particle ...

A force `F=(10+0.50x)` acts on a particle in the x direction, where F is in newton and x in meter./find the work done by this force during a displacement form x=0 tox=2.0m

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As the force is variable, we shall find the work done in a small displacement x to x+dx and then integrate it to find the total work. The work done in this small displacement is
`dW=vecF.vec(dx)=(10+0.5x)dx`
Thsu, `W=int^2.0_0(10+0.50x)dx`
`=[10x+0.50 x^2/2]^20_0=21J
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