# 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 [(pi)/12, (3pi)/8], where x is in meters?

Apr 3, 2018

The work is $= 428.1 J$

#### Explanation:

$\text{Reminder : }$

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

The work done is

$W = F \cdot d$

The frictional force is

${F}_{r} = {\mu}_{k} \cdot N$

The coefficient of kinetic friction is ${\mu}_{k} = \left(1 + 5 \cot x\right)$

The normal force is $N = m g$

The mass of the object is $m = 6 k g$

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

$= 6 \cdot \left(1 + 5 \cot x\right) g$

The work done is

$W = 6 g {\int}_{\frac{1}{12} \pi}^{\frac{3}{8} \pi} \left(1 + 5 \cot\right) \mathrm{dx}$

=6g*[x+5ln(sin(x) ] _(1/12pi)^(3/8pi)

=6g((3/8pi+5ln(sin(3/8pi))-(1/12pi+5ln(sin(1/12pi))

$= 6 \times 9.8 \times 7.28$

$= 428.1 J$