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

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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

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