Question

A body of mass $1\text{kg}$ begins to move under the action of a time dependent force $\overrightarrow{F}=(2t\hat{i}+3t^2\hat{j})\text{N}$. Calculate the instantaneous power developed at time $t$ ?

• A. $(2t^2+4t^4)\text{W}$
• B. $(2t^3+3t^4)\text{W}$
• C. $(2t^3+3t^5)\text{W}$
• D. $(2t^2+3t^3)\text{W}$

Answer 1

The correct option is C $(2t^3+3t^5)\text{W}$

Given, force
$\overrightarrow{F}=(2t\hat{i}+3t^2\hat{j})\text{N}$
So, acceleration
$\overrightarrow{a}=\frac{\overrightarrow{F}}{m}=\frac{2t\hat{i}+3t^2\hat{j}}{1}$
$\Rightarrow \overrightarrow{a}=(2t\hat{i}+3t^2\hat{j}){\text{m/s}}^2$
Velocity,
$\overrightarrow{v}=\int \overrightarrow{a}dt$
[ $\overrightarrow{a}=\frac{d\overrightarrow{v}}{dt}$]
$\overrightarrow{v}=\int (2t\hat{i}+3t^2\hat{j})dt$
$\Rightarrow \overrightarrow{v}={\int }_0^{t}2tdt\hat{i}+{\int }_0^{t}3t^{2}dt\hat{j}$
$\Rightarrow \overrightarrow{v}=(t^2\hat{i}+t^3\hat{j})\text{m/s}$
Now, power of body,
$P=\overrightarrow{F}.\overrightarrow{v}$
$\Rightarrow P=(2t\hat{i}+3t^2\hat{j}).(t^2\hat{i}+t^3\hat{j})$
$\therefore P=(2t^3+3t^5)\text{W}$