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

Engineering

Physics

Work Done by a Variable Force

Question

9 J

18 J

4.5 J

22 J

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$\frac{dv}{dt} = 6 t$

$\int_{0}^{v} dv = \int_{0}^{1} 6 t dt$

$v = {3 t^{2} |}_{0}^{1} = 3$

Work = chage in K.E

$= \frac{1}{2} \times 1 \times 9$

$= \frac{9}{2} J$