# An object with a mass of 6 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 1+cotx . How much work would it take to move the object over x in [(pi)/12, (3pi)/8], where x is in meters?

Jul 20, 2017

The work is $= 128.7 J$

#### Explanation:

We need

$\int \cot x \mathrm{dx} = \ln | \sin \left(x\right) | + C$

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 $m = 6 k g$

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

$= 6 \left(1 + \cot x\right) g N$

The work done is

$W = 6 g {\int}_{\frac{1}{12} \pi}^{\frac{3}{8} \pi} \left(1 + \cot x\right) \mathrm{dx}$

$= 6 g \cdot {\left[x + \ln | \sin \left(x\right) |\right]}_{\frac{1}{12} \pi}^{\frac{3}{8} \pi}$

=6g((3/8pi+ln|sin(3/8pi)|))-(1/12pi+ln|sin(1/12pi)|))#

$= 6 g \left(\frac{7}{24} \pi + 1.27\right)$

$= 6 g \left(2.19\right)$

$= 128.7 J$