# An object with a mass of 6 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 1+5cotx . How much work would it take to move the object over x in [(7pi)/12, (5pi)/8], where x is in meters?

Nov 27, 2017

$W = \setminus {\int}_{\frac{7 \setminus \pi}{12}}^{\frac{5 \setminus \pi}{8}} {F}_{f} \left(x\right) . \mathrm{dx} = \setminus \frac{\pi}{24} + 5 \setminus \ln \left\{\setminus \sin \frac{\frac{5 \setminus \pi}{8}}{\setminus} \sin \left(\frac{7 \setminus \pi}{12}\right)\right\}$

#### Explanation:

\mu_k(x) = 1 + 5\cotx; \qquad F_{f} (x) = \mu_k(x)mg; #

$W = \setminus {\int}_{\frac{7 \setminus \pi}{12}}^{\frac{5 \setminus \pi}{8}} {F}_{f} \left(x\right) . \mathrm{dx}$
$= \setminus {\int}_{\frac{7 \setminus \pi}{12}}^{\frac{5 \setminus \pi}{8}} \left(1 + 5 \setminus \cot x\right) \mathrm{dx} = {\left[x + 5 \left(\setminus \ln \left(| \setminus \sin x |\right)\right)\right]}_{\frac{7 \setminus \pi}{12}}^{\frac{5 \setminus \pi}{8}}$
$= \left(\frac{5 \setminus \pi}{8} - \frac{7 \setminus \pi}{12}\right) + 5 \setminus \times \left\{\setminus \ln \left[\setminus \sin \left(\frac{5 \setminus \pi}{8}\right)\right] - \setminus \ln \left[\setminus \sin \left(\frac{7 \setminus \pi}{12}\right)\right]\right\}$
$= \setminus \frac{\pi}{24} + 5 \setminus \ln \left\{\setminus \sin \frac{\frac{5 \setminus \pi}{8}}{\setminus} \sin \left(\frac{7 \setminus \pi}{12}\right)\right\}$