Question

A body of mass 3 kg is under a constant force which causes a displacement s in meters in it, given by the relation s = ($\frac{1}{3}$) t2, where t is in s. Work done by the force in 2 s is :

• A. ($\frac{17}{3}$)J
• B. ($\frac{3}{8}$)J
• C. ($\frac{8}{3}$)J
• D. ($\frac{3}{17}$)J

Answer 1

Answer: (C)

($\frac{8}{3}$)J

Answer 2

Work Done by Constant Force: The work done by a constant force is given by the formula W = F$\times$d, where F is the force and d is the displacement.
In this case, s = ($\frac{1}{3}$)t2, so the force is F = m$\times$($\frac{d^2}{dt^2}$) = $\frac{2t}{3}$.
To find the work done in 2 seconds, integrate F with respect to t over the interval [0, 2]:
W = $\int_{0}^{2}$ ($\frac{2t}{3}$) dt = ($\frac{2}{3}$)$\times$[$\frac{t^2}{2}$] from 0 to 2 = ($\frac{2}{3}$)$\times$($\frac{2^2}{2}-0$) = ($\frac{8}{3}$) J.
So, the correct option is ($\frac{8}{3}$) J.