Question

The position $x$ of a particle having mass $m$ moving along $x$ - axis at time $t$ is given by the equation $t=\sqrt{x}+2$, where $x$ is in $\text{m}$ and $t$ in $\text{s}$. Find the work done by the force in first $4\text{s}$.

• A. $0\text{J}$
• B. $2\text{J}$
• C. $4\text{J}$
• D. $8\text{J}$

Answer 1

The correct option is A $0\text{J}$

Given:
$t=\sqrt{x}+2$
$\Rightarrow (\sqrt{x}{)}^2=(t-2{)}^2$
$\Rightarrow x=(t-2{)}^2$
$\Rightarrow \frac{dx}{dt}=2(t-2)$
$\Rightarrow v=2(t-2)$
At $t=0$,
$v_i=-4\text{m/s}$
At $t=4\text{s}$,
$v_f=4\text{m/s}$
We know that, by the work energy theorem,
Work done by the force $=$ Change in kinetic energy
$w=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$
$\Rightarrow w=\frac{1}{2}m(4{)}^2-\frac{1}{2}m(-4{)}^2=0\text{J}$