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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Please answer the question in detail
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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