Question

A force $F=(10+0.5x)$ acts on a particle in the $x$ direction, where $F$ is in $\text{newton}$ and $x$ is in$\text{metre}$. Find the work done by this force during the displacement from $x=0$ to $x=2.0\text{m}$

• A. $18\text{J}$
• B. $19\text{J}$
• C. $20\text{J}$
• D. $21\text{J}$

Answer 1

The correct option is D $21\text{J}$

The force acting on the body is variable, therefore we will integrate the work done by force during small displacement $dx$ to obtain the net work done during the displacement from $x=0$ to $x=2\text{m}$.

Given, $F=(10+0.5x)$

The work done during small displacement $\left(dx\right)$ is,
$dW=Fdx$

Considering the force to be constant during small displacement $dx$
$\Rightarrow W={\int }_0^2(10+0.5x)dx$
$\text{Or},W={[10x+\frac{1}{4}{x}^2]}_0^2$
$\therefore W=(20+1)-0=21\text{J}$