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A force F is related to the position of a particle by the relation `F=(10x^(2))N` . Find the work done by the force when the particle moves from `x=2m to x=4m`.

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`W=int_(2)^(4)Fdx=int_(2)^(4) 10x^(2)dx=[(10x^(3))/(3)]_(2)^(4)`
`=(10)/(3)[4^(3)-2^(3)]=(10)/(3)xx56=186.7J`
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