Question

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

• A. $42J$
• B. $38J$
• C. $27J$
• D. $21J$

Answer 1

The correct option is D $21J$

Small amount of work done dW in giving a small displacement $d\overrightarrow{x}$ is given by
$dW=\overrightarrow{x}$
or $dW=Fdxcos{0}^{\circ }$
or $dW=Fdx[\therefore cos{0}^{\circ }=1]$
Total work done, W = ${\int }_{x=0}^{x=2}Fdx={\int }_{x=0}^{x=2}$ $(0.5x+10)dx$
$={\int }_{x=0}^{x=2}0.5xdx+{\int }_{x=0}^{x=2}10dx=0.5{\left|\frac{x^2}{2}\right|}_{x=0}^{x=2}+10{\left|x\right|}_{x=0}^{x=2}$
$=\frac{0.5}{2}[2^2-0^2]+10(2-0]=(1+20)=21J$