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 .

Answer 1

Here, `m=2kg, x=t^(3)//3, W=?t=2s,`
`upsilon=(dx)/(dt)=(3t^(3))/(3)=t^(2),a=(dupsilon)/(dt)=2t`
`F=ma=2xx2t=4t`
`W=int_(0)^(x)Fdx=int_(0)^(2)4t.t^(2)dt=4[(t^(4))/(4)]_(0)^(2)=16J`