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

Answer 1

The correct option is A

$4.5J$

Step1: Given data.

Force, $F=6t$

Mass, $m=1kg$

Step2: Find the work done by the force during the first $1sec$

We know that,

$F=ma$

Where, $F$ is force, $m$ is mass and $a$ is acceleration.

$\Rightarrow$ $F=m\times \frac{dv}{dt}$ $\left[\because a=\frac{dv}{dt}\right]$

$\Rightarrow$ $6t=1\times \frac{dv}{dt}$

$\Rightarrow$ $dv=6t\times dt$ ….$\left(i\right)$

Integrating the equation $\left(i\right)$ both the sides with velocity $0$ to $v$ and time $0$ to $1$ we get,

$\Rightarrow$${\int }_0^{v}dv={\int }_0^{1}6t\times dt$

$\Rightarrow$ $v=6\times {\left[\frac{t^2}{2}\right]}_0^1$

$\Rightarrow$ $v=3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ ….$\left(ii\right)$

Now,

Also, we know that

$W=∆KE=\frac{1}{2}m{v}^2$

Where, $W$ is the work done.

$\Rightarrow$$W=\frac{1}{2}\times 1\times 3^2$

$\Rightarrow$$W=4.5J$

Hence, option A is correct.