Q 11-05-196JEE MainJEE Main 2017 (2 Apr)Easy
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 $1$ s will be:
Answer: (B) $4.5$ J
$a = F/m = 6t$, so with $v(0)=0$:
$$v = \int_0^t 6t\,dt = 3t^2 \;\Rightarrow\; v(1) = 3\ \text{m/s}$$
By the work–energy theorem,
$$W = \tfrac12 mv^2 = \tfrac12(1)(3)^2 = 4.5\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics