Under the action of a force, a `2 kg` body moves such that its position x as a function of time is given by `x =(t^(3))/(3)` where x is in metre and t in second. The work done by the force in the first two seconds is .

Options

A. `1.6 J`
B. `16 J`
C. `160 J`
D. `1600 J`

Answer

B

Explanation

`v = (dx)/(dt) = (d)/(dt) ((t^(3))/(3)) = t^(2)`
Where `t = 0` When `t = 2`, then `t = 4 m//s`
Work down in first two second = change in KE
`W = (1)/(2)m [(4)^(2) - (0)^(2)] = (1)/(2) 2 xx 16 = 16`