Question

A Force $\overrightarrow{F}=(6\hat{i}+4\hat{j}-2\hat{k})\text{N}$ is applied on a block which is moving with velocity $\overrightarrow{v}=(8\hat{i}+12\hat{j}-3\hat{k})\text{m/s}$. The instantaneous power applied to the particle is

• A. $96\text{W}$
• B. $102\text{W}$
• C. $108\text{W}$
• D. $90\text{W}$

Answer 1

The correct option is B $102\text{W}$

Given:
Force, $\overrightarrow{F}=(6\hat{i}+4\hat{j}-2\hat{k})\text{N}$
Velocity, $\overrightarrow{v}=(8\hat{i}+12\hat{j}-3\hat{k})\text{m/s}$
We know power delivered by a force is given by,
$P=\overrightarrow{F}.\overrightarrow{v}$, where $\overrightarrow{F}$ is force applied and $\overrightarrow{v}$ is velocity of block.
$P=(6\hat{i}+4\hat{j}-2\hat{k}).(8\hat{i}+12\hat{j}-3\hat{k})$
$\Rightarrow P=6\times 8+4\times 12+[-2\times (-3\left)\right]$
$\Rightarrow P=48+48+6$
$\Rightarrow P=102\text{W}$
Hence option $\left(b\right)$ is correct.