Question

A time dependent force `F=6t` acts on a particle of mass `1kg`. If the particle starts from rest, the work done by the force during the first `1 sec.` will be
A. `9J`
B. `18 J`
C. `4.5 J`
D. `22 J`

Answer 1

Here, `F=6t, m=1kg, u=0, W=?`
Now, `F=ma=m(dupsilon)/(dt)=1xx(dupsilon)/( dt)`
`:. (dupsilon)/(dt)=F=6t`
`dupsilon=6t dt`
`int_(0)^(upsilon)d upsilon=int_(0)^(1)6t dt`
`upsilon=[6(t^(2))/(2)]_(0)^(1)=3(1^(2)-0)=3m//s`
From work energy theorem,
`W=DeltaKE=(1)/(2)m(upsilon^(2)-u^(2))`
`=(1)/(2)xx1(3^(2)-0^(2))=4.5J`