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,

$P=\overrightarrow{F}.\overrightarrow{v}$

$\Rightarrow P=(2t\hat{i}+3t^2\hat{j}).(t^2\hat{i}+t^3\hat{j})$

$\therefore P=(2t^3+3t^5)\text{W}$