In the given figure F=10N , R=1 m , mass of the body is 2 kg and moment of inertia of the body about an axis passing through O and perpendicular to plane of body is `4 kg-m^(2)` .O is the centre of mass of the body.

If ground is sufficiently rough , what is kinetic energy of the body now in the given time interval ?

Options

A. `18.75J`
B. `16.67J`
C. `5.55J`
D. cannot be found

Answer

B

Explanation

`tau=tau_A/I_A=(F4R-F3R)/(I_0+m(2R)^2)`

`tau=(FR)/(4+2(2x1)^2)=(10xx1)/12=5/6rad//s^2`
After `2s: omega=alphat=5/6xx2=5/3rad//s`
`KE=1/2mv^2+1/2I_0(omega^2)`
`=1/2xx2(omega2R)^2+1/2 4(omega)^2`
`=1/2xx2xx(5/3xx2xx2)^2+1/2xx4(5/3)^2`
`=150/9=16.67J`