Question

A body of mass $2\text{kg}$ is travelling with uniform velocity from $(1,2,4)\text{m}$ to $(3,4,6)\text{m}$ in $4\text{seconds}$. The kinetic energy of the body is

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

Answer 1

The correct option is B $\frac{3}{4}\text{J}$

Position vector $\overrightarrow{S}$ is given by
$\overrightarrow{S}=(3-1)\hat{i}+(4-2)\hat{j}+(6-4)\hat{k}$
$=2\hat{i}+2\hat{j}+2\hat{k}$
Velocity vector $\overrightarrow{v}$ is given by the dividing the position vector by time
$\overrightarrow{v}=\frac{\overrightarrow{S}}{t}=\frac{2\hat{i}+2\hat{j}+2\hat{k}}{4}=\left(\frac{\hat{i}+\hat{j}+\hat{k}}{2}\right)$

We know that kinetic energy is given by,
$=\frac{1}{2}m(\overrightarrow{v}.\overrightarrow{v})$

$\Rightarrow \overrightarrow{v}.\overrightarrow{v}=\frac{1}{4}(\hat{i}+\hat{j}+\hat{k}).(\hat{i}+\hat{j}+\hat{k})$

$\Rightarrow \frac{1}{2}m(\overrightarrow{v}.\overrightarrow{v})=\frac{1}{2}\times 2\times \left(\frac{1}{4}\right(1\times 1+1\times 1+1\times 1))$
Kinetic Energy $=\frac{3}{4}\text{J}$

Hence option B is the correct answer.