Question 154: Work, Energy and Power & Circular Motion

Question

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:

Accepted Answer

Given that, F = 6t
We know, F = ma = $$m{{dv} \over {dt}}$$
$$	herefore$$ $$m{{dv} \over {dt}} = 6t$$
$$ \Rightarrow $$ $$1.{{dv} \over {dt}} = 6t$$ [as m = 1]
$$ \Rightarrow $$ $$\int\limits_0^v {dv}  = \int {6t} dt$$
$$ \Rightarrow $$ $$v = 6\left[ {{{{t^2}} \over 2}} ight]_0^1$$
$$ \Rightarrow $$ $$v = {6 \over 2} = 3$$ m/s [ as given t = 1 sec ]
Work done by the body during the first 1 form work-energy theorem,
W = $$\Delta $$K.E = $${1 \over 2}m\left( {{V^2} - {v^2}} ight)$$
= $${1 \over 2}.1.\left( {{3^2} - {0^2}} ight)$$ = 4.5 J

Answer

18 J

Answer

4.5 J

Answer

22 J

Answer

9 J

Work, Energy and Power & Circular Motion

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:

Choose the correct answer: