# An object with a mass of 8 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 2x^2-3x . How much work would it take to move the object over #x in [2, 3], where x is in meters?

Mar 20, 2016

$797.1 J$ rounded to one place of decimal

#### Explanation:

Force of kinetic friction which needs to be overcome to move the object

${F}_{k} =$Coefficient of kinetic friction ${\mu}_{k} \times$normal force $\eta$
where $\eta = m g$
In the problem ${\mu}_{k}$ is stated to be ${u}_{k} \left(x\right)$
Inserting given quantities and taking the value of $g = 9.8 m / {s}^{2}$
${F}_{k} = \left(2 {x}^{2} - 3 x\right) \times 8 \times 9.8 N$
${F}_{k} = 78.4 \left(2 {x}^{2} - 3 x\right) N$

When this force moves through a small distance $\mathrm{dx}$, the work done is given as
${F}_{k} \cdot \mathrm{dx} = \left[78.4 \left(2 {x}^{2} - 3 x\right)\right] \cdot \mathrm{dx}$
When the force moves through a distance from $x \in \left[2 , 3\right]$, total work done is integral of RHS over the given interval.

Total work done$= {\int}_{2}^{3} 78.4 \left(2 {x}^{2} - 3 x\right) \cdot \mathrm{dx}$
$\implies$Total work done$= 78.4 {\int}_{2}^{3} \left(2 {x}^{2} - 3 x\right) \cdot \mathrm{dx}$

$= 78.4 \left(2 {x}^{3} / 3 - 3 {x}^{2} / 2 + C\right) {|}_{2}^{3}$, where C is constant of integration.
$= 78.4 \left[\left(2 {3}^{3} / 3 - 3 {3}^{2} / 2 + \cancel{C}\right) - \left(2 {2}^{3} / 3 - 3 {2}^{2} / 2 + \cancel{C}\right)\right]$
$= 78.4 \left[18 - \frac{27}{2} - \frac{16}{3} + 6\right]$
$= 78.4 \left[\frac{108 - 51 - 32 + 36}{6}\right]$
$= 78.4 \left[\frac{61}{6}\right]$
$797.1 J$ rounded to one place of decimal