# An object with a mass of 7 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 1+xcos(pi/4-x/6) . How much work would it take to move the object over #x in [0, 4pi], where x is in meters?

Jul 13, 2017

The work is $= 494.6 J$

#### Explanation:

The integration by parts is

$\int u v ' \mathrm{dx} = u v - \int u ' v \mathrm{dx}$

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

$\mathrm{dW} = {F}_{r} \cdot \mathrm{dx}$

We start by calculating the integral of

$\int x \cos \left(\frac{\pi}{4} - \frac{x}{6}\right)$

We perform this by integration by parts

$u = x$, $\implies$, $u ' = 1$

$v ' = \cos \left(\frac{\pi}{4} - \frac{x}{6}\right)$, $\implies$, $v = - 6 \sin \left(\frac{\pi}{4} - \frac{x}{6}\right)$

Therefore,

$\int x \cos \left(\frac{\pi}{4} - \frac{x}{6}\right) = - 6 x \sin \left(\frac{\pi}{4} - \frac{x}{6}\right) + \int 1 \cdot 6 \sin \left(\frac{\pi}{4} - \frac{x}{6}\right)$

$= - 6 x \sin \left(\frac{\pi}{4} - \frac{x}{6}\right) + 36 \cos \left(\frac{\pi}{4} - \frac{x}{6}\right)$

The work is

$W = 7 g {\int}_{0}^{4 \pi} \left(1 + x \sin \left(\frac{\pi}{4} - \frac{x}{6}\right)\right)$

$= 7 g {\left[x - 6 x \sin \left(\frac{\pi}{4} - \frac{x}{6}\right) + 36 \cos \left(\frac{\pi}{4} - \frac{x}{6}\right)\right]}_{0}^{4 \pi}$

$= 7 g \left(\left(4 \pi - 24 \pi \sin \left(- \frac{5}{12} \pi\right) + 36 \cos \left(- \frac{5}{12} \pi\right)\right) - \left(0 - 0 + 36 \cos \left(\frac{\pi}{4}\right)\right)\right)$

$= 7 g \left(69.259\right)$

$= 494.6 J$