Question

A force $F=(2x+3x^2)\text{N}$ acts on a particle in $x-$ direction, where $F$ is in Newton and $x$ is in metre. Find the work done by this force for the displacement from $x=0\text{m}$ to $x=3\text{m}$.

• A. $20\text{J}$
• B. $36\text{J}$
• C. $45\text{J}$
• D. $56\text{J}$

Answer 1

The correct option is B $36\text{J}$

Given:
$F=(2x+3x^2)\text{m}$

Amount of work done in moving a particle through small distance $dx$ is given below

$dW=\overrightarrow{F}.\overrightarrow{dx}$

Since, force $F$ and displacement $dx$ have same direction

$\Rightarrow dW=(2x+3x^2)dx$

We can obtain the total work done from $x=0$ to $x=3\text{m}$ by integrating this as follows.

$W=\underset{x=0}{\overset{x=3}{\int }}(2x+3x^2)dx$

Since, $\int x^{m}dx=\frac{x^{m+1}}{m+1}$

$\therefore W={\{\frac{2x^2}{2}+\frac{3x^3}{3}\}}_0^3$

$\Rightarrow W={\begin{array}{c}{x}^2+x^3\end{array}|}_0^3=3^2+3^3=36\text{J}$

So, option $\left(b\right)$ is correct.