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(pi/4-x/6) . How much work would it take to move the object over #x in [0, 4pi], where x is in meters?

Nov 16, 2017

The work is $= 10166.5 J$

Explanation:

We need

$\int x \cos \left(\frac{\pi}{4} - \frac{x}{6}\right) \mathrm{dx} = 18 \sqrt{2} \left(\frac{x}{6} \sin \left(\frac{x}{6}\right) - \frac{x}{6} \cos \left(\frac{x}{6}\right) + \cos \left(\frac{x}{6}\right) + \sin \left(\frac{x}{6}\right)\right) + C$

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 $m = 7 k g$

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

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

The work done is

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

$= 7 g \cdot {\left[x + {x}^{2} / 2 + 18 \sqrt{2} \left(\frac{x}{6} \sin \left(\frac{x}{6}\right) - \frac{x}{6} \cos \left(\frac{x}{6}\right) + \cos \left(\frac{x}{6}\right) + \sin \left(\frac{x}{6}\right)\right)\right]}_{0}^{4 \pi}$

$= 7 g \left(148.2\right)$

$= 10166.5 J$