Question

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

Answer 1

Correct Answer - C
`v=(dx)/(dt) =t^(2)`
`W =1/2mv^(2) =1/2mt^(4)`
`1/2 xx 2 xx (2)^(4) =16 J`