# An object with a mass of 6 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 1+3cscx . How much work would it take to move the object over #x in [(pi)/6, (pi)/4], where x is in meters?

Apr 1, 2016

$w = {\int}_{\frac{\pi}{6}}^{\frac{\pi}{4}} {\mu}_{k} \left(x\right) \cdot m \cdot g \mathrm{dx}$
$w = m g {\int}_{\frac{\pi}{6}}^{\frac{\pi}{4}} \left(1 + 3 \csc x\right) \mathrm{dx}$
$w = m g {\int}_{\frac{\pi}{6}}^{\frac{\pi}{4}} \mathrm{dx} + 3 m g {\int}_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc x \mathrm{dx}$

$= m g {\int}_{\frac{\pi}{6}}^{\frac{\pi}{4}} \mathrm{dx} + 3 m g {\int}_{\frac{\pi}{6}}^{\frac{\pi}{4}} \csc x \mathrm{dx}$

$= m g \left[\left(\frac{\pi}{4} - \frac{\pi}{6}\right)\right] \left(\frac{\pi}{6}\right) - 3 m g \ln \left(\csc \left(\frac{\pi}{4}\right) + \cot \left(\frac{\pi}{4}\right) - \ln \left(\csc \left(\frac{\pi}{6}\right) + \cot \left(\frac{\pi}{6}\right)\right)\right)$
pl put m=6kg
$g = 9.8 m {s}^{-} 2$
and values from trigonometrical table and calculate
$w = 6 \times 9.8 \cdot \frac{\pi}{12} - 3 \times 6 \times 9.8 \times \ln \left(\frac{\sqrt{2} + 1}{2 + \sqrt{3}}\right) J$