# An object with a mass of 6 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 2+cscx . How much work would it take to move the object over #x in [(3pi)/4, (7pi)/8], where x is in meters?

Jul 29, 2017

The work is $= 89.3 J$

#### Explanation:

We need

$\int \csc x \mathrm{dx} = \ln | \tan \left(\frac{x}{2}\right) |$

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 $m = 6 k g$

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

$= 6 \left(2 + \csc x\right) g$

The work done is

$W = 6 g {\int}_{\frac{3}{4} \pi}^{\frac{7}{8} \pi} \left(2 + \csc x\right) \mathrm{dx}$

$= 6 g \cdot {\left[2 x + \ln | \tan \left(\frac{x}{2}\right) |\right]}_{\frac{3}{4} \pi}^{\frac{7}{8} \pi}$

$= 6 g \left(\left(2 \cdot \frac{7}{8} \pi + \ln | \tan \left(\frac{7}{16} \pi\right) |\right) - \left(2 \cdot \frac{3}{4} \pi + \ln | \tan \left(\frac{3}{8} \pi\right) |\right)\right)$

$= 6 g \left(\frac{7}{4} \pi - \frac{6}{4} \pi + 1.61\right)$

$= 6 g \left(\frac{1}{4} \pi + 0.73\right)$

$= 89.3 J$