# The force applied against an object moving horizontally on a linear path is described by F(x)=x^2-x+1 . By how much does the object's kinetic energy change as the object moves from  x in [ 1, 2 ]?

Feb 8, 2018

$\frac{11}{6} J$

#### Explanation:

Suppose,initially the object had a velocity of $u$ after going from $x = 1 \to x = 2$ its velocity became $v$,so change in kinetic energy = $\frac{1}{2} m {v}^{2} - \frac{1}{2} m \cdot {u}^{2}$

Given,

$F = {x}^{2} - x + 1$

Or, m×a=x^2-x+1

Or, $m v \frac{\mathrm{dv}}{\mathrm{dx}} = {x}^{2} - x + 1$

Or, $m v \mathrm{dv} = {x}^{2} \mathrm{dx} - x \mathrm{dx} + \mathrm{dx}$

Or, $m {\int}_{u}^{v} \mathrm{dv} = {\int}_{1}^{2} {x}^{2} \mathrm{dx} - {\int}_{1}^{2} x \mathrm{dx} + {\int}_{1}^{2} \mathrm{dx}$

Or, $\frac{m {v}^{2}}{2} - \frac{m \cdot {u}^{2}}{2} = \frac{11}{6}$