# An object with a mass of 7 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 1-cos(pi/4-x/6) . How much work would it take to move the object over #x in [0, 8pi], where x is in meters?

Nov 26, 2017

The work is $= 1539.6 J$

#### Explanation:

We need to calculate the integral $\int \cos \left(\frac{1}{4} \pi - \frac{x}{6}\right) \mathrm{dx}$

Perform this integration by substitution

Let $u = \frac{1}{4} \pi - \frac{x}{6}$, $\implies$, $\mathrm{du} = - \frac{1}{6} \mathrm{dx}$

$\int \cos \left(\frac{1}{4} \pi - \frac{x}{6}\right) \mathrm{dx} = - 6 \int \cos u \mathrm{du}$

$= - 6 \sin u = - 6 \sin \left(\frac{1}{4} \pi - \frac{x}{6}\right)$

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 $g = 9.8 m {s}^{-} 2$

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

$= 7 \cdot \left(1 - \cos \left(\frac{\pi}{4} - \frac{x}{6}\right)\right) g$

The work done is

$W = 7 g {\int}_{0}^{8 \pi} \left(1 - \cos \left(\frac{\pi}{4} - \frac{x}{6}\right)\right) \mathrm{dx}$

$= 7 g {\left[x + 6 \sin \left(\frac{\pi}{4} - \frac{x}{6}\right)\right]}_{0}^{8 \pi}$

$= 7 g \left(8 \pi + 6 \sin \left(\frac{\pi}{4} - \frac{8}{6} \pi\right)\right) - \left(6 \sin \left(\frac{\pi}{4}\right)\right)$

$= \left(7 \times g \times 22.44\right) J$

$= 1539.6 J$