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

Nov 25, 2017

The work is $= 5501.7 J$

#### Explanation:

We need to calculate the integral $\int x \cos \left(\frac{x}{6}\right) \mathrm{dx}$

Perform this integration by parts

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

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

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

$\int x \cos \left(\frac{x}{6}\right) \mathrm{dx} = 6 x \sin \left(\frac{x}{6}\right) - 6 \int \sin \left(\frac{x}{6}\right) \mathrm{dx}$

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

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 $m = 7 k g$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

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

$= 7 \cdot \left(1 + x - x \cos \left(\frac{x}{6}\right)\right) g$

The work done is

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

$= 7 g {\left[x + \frac{1}{2} {x}^{2} - 6 x \sin \left(\frac{x}{6}\right) - 36 \cos \left(\frac{x}{6}\right)\right]}_{0}^{4 \pi}$

$= 7 g \left(\left(4 \pi + \frac{1}{2} {\left(4 \pi\right)}^{2} - 24 \pi \sin \left(\frac{2}{3} \pi\right) - 36 \cos \left(\frac{2}{3} \pi\right)\right) - \left(0 + 0 - 0 - 36\right)\right)$

$= 7 g \left(80.2\right)$

$= 5501.7 J$