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

Options

A. 9 J
B. 18 J
C. 4.5 J
D. 22 J

Answer

C

Explanation

Using ,F=ma=`m(dV)/(dt)`
6t=1.`(dV)/(dt)`[`because` m=1 kg given]
`underset(0)overset(v)int dV=int 6t dtimpliesV=6[(t^(2))/(2)]_(0)^(t)=3m^(-1)[t=1sec]`
From work -energy theorem,
W=`DeltaKE=(1)/(2)m(V^(2)-u^(2))=(1)/(2)xx1xx9=4.5 J`