Question

A time dependent force $F=6t$ acts on a particle of mass $1kg$. If the particle starts from rest, the work done by the force during the first $1sec$. will be :

• A. 4.5 J
• B. 22 J
• C. 9 J
• D. 18 J

Answer 1

Answer: (A)

Given,
$F=6t$
$m=1kg$
Force, $F=ma=6t$
$\begin{array}{l}1\times \frac{dv}{dt}=6t\\ dv=6tdt\end{array}$
Integrating both side,
$\begin{array}{l}v=\int dv={\int }_0^{1}6tdt\\ v=6{\left[\frac{t^2}{2}\right]}_0^1\\ v=3m/s\end{array}$
Initial velocity, $u=0m/s$
From work energy theorem,
$\begin{array}{rl}W& =K_f-K_i\\ W& =\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2\\ W& =\frac{1}{2}\times 1\times 3\times 3-\frac{1}{2}\times 1\times 0\times 0\\ W& =4.5-0=4.5J\end{array}$