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

Oct 9, 2017

The work is $= 180.9 J$

Explanation:

We need to compute

$I = \int x \ln x \mathrm{dx}$

Apply the integration by parts

$u = \ln x$, $\implies$, $u ' = \frac{1}{x}$

$v ' = x$, $\implies$, $v = {x}^{2} / 2$

$I = u v - \int u ' v \mathrm{dx} = \frac{1}{2} {x}^{2} \ln x - \int \frac{1}{x} \cdot {x}^{2} / 2 \mathrm{dx}$

$= \frac{1}{2} {x}^{2} \ln x - \frac{1}{2} \int x \mathrm{dx} = \frac{1}{2} {x}^{2} \ln x - \frac{1}{2} \cdot {x}^{2} / 2$

$= \frac{1}{2} {x}^{2} \ln x - \frac{1}{4} {x}^{2} + C$

The work done is

$W = F \cdot d$

The frictional force is

${F}_{r} = {\mu}_{k} \cdot N$

The normal force is $N = m g$

The mass is $m = 8 k g$

${F}_{r} = {\mu}_{k} \cdot m g$

$= 8 \cdot \left(x \ln x\right) g$

The work done is

$W = 8 g {\int}_{2}^{3} \left(x \ln x\right) \mathrm{dx}$

$= 8 g \cdot {\left[\frac{1}{2} {x}^{2} \ln x - \frac{1}{4} {x}^{2}\right]}_{2}^{3}$

$= 8 g \left(\frac{1}{2} \cdot 9 \cdot \ln 3 - \frac{1}{4} \cdot 9\right) - \left(\frac{1}{2} \cdot 4 \ln 2 - \frac{1}{4} \cdot 4\right)$

$= 8 g \left(\frac{9}{2} \ln 3 - 2 \ln 2 - \frac{5}{4}\right)$

$= 180.9 J$