Under the action of a constant force, a 2 kg body moves such that its position along X-axis is given by `x=(t^(2))/(3)` where x is in metres and in seconds and x is function time. The work done in 2 sec is:

Options

A. 1600 J
B. 169 J
C. 16 J
D. 1.6 J

Answer

c

Explanation

`x=(t^(3))/(3)implies(dv)/(dt)=r^(2)impliesdx=r^(2)dt`
Also `a=(d^(2)x)/(dt^(2))=2r`.
Now F=`ma=2xx2t=`4t.
`dW=F dx=4r.t^(2) dt=4 r^(3) dt`.
`W=int_(0)^(2)4t^(3)dt=[(4t^(4))/(4)]_(0)^(2)=16 J`.