# The force applied against an object moving horizontally on a linear path is described by F(x)= x^2 + x N . By how much does the object's kinetic energy change as the object moves from  x in [ 2, 5 ]?

Jul 14, 2016

$- 49.5 J$

#### Explanation:

The change in KE is work done against varible frictional force.

So

$- {\int}_{2}^{5} F \left(x\right) \mathrm{dx}$

$= - {\int}_{2}^{5} \left({x}^{2} + x\right) \mathrm{dx}$

$= - {\left[{x}^{3} / 3 + {x}^{2} / 2\right]}_{2}^{5}$

$= - \left({5}^{3} / 3 + {5}^{2} / 2 - {2}^{3} / 3 - {2}^{2} / 2\right)$

$= - \left(\frac{117}{3} + \frac{21}{2}\right) J$

$= - \left(39 + 10.5\right) = - 49.5 J$

$\textcolor{red}{\text{Negative sign means work done against the force}}$

$\textcolor{red}{\text{,representing decrease in KE}}$