A particle of mass m is subjected to a force `vecF=F_(0)[cos (t)hati+sin(1)hatj]`. If initially (t=0) the particle was at rest, the kinetic energy of the particle as a function of time is given by:

Options

A. `(F_(0)^(2))/(m)(1-cos(2t))`
B. `(F_(0)^(2))/(m)(1-cost)`
C. `(F_(0)^(2))/(m) sin(t)`
D. `(F_(0)^(2))/(m)t`

Explanation

`veca=(vecF)/(m)=(F_(0))/(m) [cos(t)hati+sin(t)hatj]`
`vecv=int_(0)^(t)vecadt=(F_(0))/(m)[sin(t)hati+(1-cos t)hatj]`
`KE=(1)/(2)m(vecv.vecv)=(1)/(2)m[(F_(0)^(2))/(m^(2))(2-2cos t)]`
`=(F_(0)^(2))/(m) (1-cos t)`