Question

A body of mass $2\text{kg}$ starts from rest, uniformly accelerates from the point $(1,2,3)$ to point $(2,3,5)$ in $3$ seconds. The net work done by the body in $3$ seconds is

• A. $\frac{3}{4}\text{J}$
• B. $\frac{2}{3}\text{J}$
• C. $\frac{4}{9}\text{J}$
• D. $\frac{9}{16}\text{J}$

Answer 1

The correct option is B $\frac{2}{3}\text{J}$

Displacement vector $\overrightarrow{S}$ is given by
$\overrightarrow{S}=(2-1)\hat{i}+(3-2)\hat{j}+(5-3)\hat{k}$
$=\hat{i}+\hat{j}+2\hat{k}$
Velocity vector $\overrightarrow{v}$ obtained by dividing displacement vector by the time taken.
So, final velocity $\overrightarrow{v}=\frac{\hat{i}+\hat{j}+2\hat{k}}{3}$
Initial velocity $=0$
Now,
According to work energy theorem
Net work done = Change in K.E
Net work done $=\frac{1}{2}m(\overrightarrow{v}.\overrightarrow{v})$
$\Rightarrow \frac{1}{2}\times 2\times (\frac{1}{9}\times (1\times 1+1\times 1+2\times 2))$
$=\frac{6}{9}\text{J}=\frac{2}{3}\text{J}$