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A particale moves under the effect of a ...

A particale moves under the effect of a force `F = Cs` from `x = 0` to `x = x_(1)`. The work down in the process is

A

`Cx_(1)^(2)`

B

`(1)/(2)Cx_(1)^(2)`

C

`Cx_(1)`

D

Zero

Text Solution

Verified by Experts

The correct Answer is:
B
`W int_(0)^(x_(1))F.dx=int_(0)^(x_(1))Cx dx =C[(x^(2))/(2)]_(0)^(x_(1))=(1)/(2)Cx_(1)^(2)`
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