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 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(12– 0) = 3m/s

From work-energy theorem

W =Δ KE = ½ m(v2 – u2)

W = ½ x 1(32 – 0) = 4.5 J

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