Question

A force $F=0.5x+10$ acts on a particle in the x direction. Here $F$ is in newton and $x$ is in metre. Calculate the work done by the force during the displacement of the particle from $x=0$ to $x=2$ metre.

• A. 9 J
• B. 21 J
• C. 15 J
• D. 36 J

Answer 1

The correct option is B

21 J

$W=\overrightarrow{F}.\overrightarrow{x}=Fxcos\theta$

$\therefore dW=\overrightarrow{F}.\overrightarrow{dx}=Fdxcos\theta$

A force is along the +ve $x$ direction, $\theta =0^{\circ }$

$\therefore dW=Fdx$

$\Rightarrow \underset{0}{\overset{w}{\int }}dw=\underset{0}{\overset{2}{\int }}Fdx$

$\Rightarrow w=\underset{0}{\overset{2}{\int }}(0.5x+10)dx$

= $0.5X\frac{x^2}{2}+10x{|}_0^2$

= $\frac{1}{2}X\frac{4}{2}+20$

= $21J$