Under the action of a force, a 3 kg body moves such that `x= (t^2)/(2)`, where position x is in metre and t is in second. What is the work done by the force in first 3 second?

Explanation

`x =(t^2)/2`
`v = (dx)/(dt) = (2t)/2 = t, ((dx)/(dt))_(t=0) = 0 and ((dx)/(dt))_(t=3) = 3 ms^(-1)`
`W = Delta k = 1/2 m(v^2 - u^2) = 1/2 xx 3 xx [3^2 - 0^2] = 13.5 J`.