Question

A force $F=10+2y$ acts on a particle in $y$ - direction, where $F$ is in Newtons and $y$ is in metres. Work done by this force to move the particle from origin to $2\text{m}$ along $+y$ direction is

• A. $24\text{J}$
• B. $32\text{J}$
• C. $48\text{J}$
• D. $12\text{J}$

Answer 1

The correct option is A $24\text{J}$

Given, variable force $\left(F\right)=10+2y$
Workdone by variable force, $W={\int }_{y_1}^{y_2}F.dy={\int }_0^2(10+2y)dy$
$=10y{|}_0^2+\frac{2y^2}{2}{|}_0^2$

$=10y{|}_0^2+y^2{|}_0^2$

$=20+4=24\text{J}$