The force applied against a moving object travelling on a linear path is given by F(x)=x+2e^x . How much work would it take to move the object over x in [0,2 ] ?

Jan 17, 2016

Work Done $= 2 {e}^{2} \mathmr{and} \approx 14.78 J$

Explanation:

Work done by a Force $\vec{F}$ when it moves the object through a distance $\vec{\mathrm{dS}}$ is

$\mathrm{dW} = \vec{F} . \vec{\mathrm{dS}} = | \vec{F} | . | \vec{\mathrm{dS}} | . \cos \theta$

Where $\theta$ is the angle between the two vectors

As the Force is being applied on the object moving in a linear path the angle between the two is $0$, an therefore $\cos \theta = 1$

$\mathrm{dW} = | \vec{F} | . | \vec{\mathrm{dS}} | .$

When the force $F \left(x\right)$ moves through a small distance $\mathrm{dx}$
$\mathrm{dW} = F \left(x\right) \mathrm{dx}$
Total work done to move the object from $x \in \left[0 , 2\right]$ is integral of $\mathrm{dW}$over limits $0$ to $2$.

$W = {\int}_{0}^{2} \left(x + 2 {e}^{x}\right) \mathrm{dx}$
$= {x}^{2} / 2 + 2 {e}^{x} + C {|}_{0}^{2}$, where C is constant of integration.
$= \left({2}^{2} / 2 + 2 {e}^{2} + C\right) - \left({0}^{2} / 2 + 2 {e}^{0} + C\right)$
=(cancel (color (red)2)+2e^2+cancel C)-((cancel (color (red)2)+cancelC)
$= 2 {e}^{2}$