Question

A particle gets displaced by$\overrightarrow{\Delta r}=(2\hat{i}+3\hat{j}+4\hat{k})m$ under the action of a force $\overrightarrow{F}=(7\hat{i}+4\hat{j}+3\hat{k})N$. The change in its kinetic energy is

• A. $38J$
• B. $70J$
• C. $52.5J$
• D. $1.26J$

Answer 1

The correct option is A

$38J$

Step 1: Given data

Displacement of the particle, $\overrightarrow{\Delta r}=(2\hat{i}+3\hat{j}+4\hat{k})m$

Force, $\overrightarrow{F}=(7\hat{i}+4\hat{j}+3\hat{k})N$

Step 2: Formula used

Change in kinetic energy,$∆KE=W$

$W=\overrightarrow{F}·\overrightarrow{∆r}$

$W$ is the work done

$\overrightarrow{F}$ is the action of force.

$\overrightarrow{∆r}$ is the displacement of the particle

Step 3: Calculate the change in Kinetic energy,

According to the work-energy theorem,

Change in kinetic energy = Work done

$∆KE=W$

We know work is done,

$$\begin{gathered} W=\overrightarrow{F}·\overrightarrow{∆r} \\ \Rightarrow W=\overrightarrow{F}·\overrightarrow{∆r} \\ \Rightarrow W=\left(7\hat{i}+4\hat{j}+3\hat{k}\right)·\left(2\hat{i}+3\hat{j}+4\hat{k}\right) \\ \Rightarrow W=14+12+12 \\ \Rightarrow W=38J \end{gathered}$$

Change in kinetic energy = Work done = $38J$

Therefore, the change in kinetic energy is $38J$.

Hence, option A is correct.