A particle is moving along the x-axis and force acting on it is given by `F=F_0sinomega_xN`, where `omega` is a constant. The work done by the force from `x=0` to `x=2` will be

Options

A. `(F_0)/(omega(1-cosomega))`
B. `(F_0)/(2omega(1-cos2omega))`
C. `(2F_0sin^2omega)/(omega)`
D. `(F_0)/(omega(1+cos2omega))`

Answer

C

Explanation

`W=int_(x1)^(x2)Fdx=int_(0)^(2)F_0sinomegaxdx`
`=F_0(|-cosomegax|_0^2)/(omega)=(F_0)/(omega)[(-cos2omega)-(-cos0)]`
`=(F_0)/(omega)(1-cos2omega)=(F_0)/(omega)[1-(1-2sin^2omega)]`
`=(2F_0sin^2omega)/(omega)`