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A particale moves under the effect of a ...

A particale moves under the effect of a force `F = Cx` 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
`Wint_0^(x1)Fdx=int_0^(x1)Cx=C[(x^2)/(2)]_0^(x1)=(1)/(2)Cx_1^2`
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