A time-dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be :
(a) 9 J
(b) 18 J
(c) 4.5 J
(d) 22 J
The correct option is c) 4.5 J
F= 6t, m = 1Kg, u =0
Now, F=ma =m(dv/dt)=1x(dv/dt)
dv/dt = 6t
\(\int ^v_0dv =\int^1_06tdt\)
v= 3(1^{2}– 0) = 3m/s
From work-energy theorem
W =Δ KE = ½ m(v^{2} – u^{2})
W = ½ x 1(3^{2} – 0) = 4.5 J