# 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 [pi/8, (3pi)/4], where x is in meters?

Jan 18, 2018

The work is $= 377.5 J$

#### Explanation:

$\text{Reminder : }$

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

The work done is

$W = F \cdot d$

The frictional force is

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

The coefficient of kinetic friction is ${\mu}_{k} = \left(2 + \csc \left(x\right)\right)$

The normal force is $N = m g$

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

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

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

The work done is

$W = 6 g {\int}_{\frac{1}{8} \pi}^{\frac{3}{4} \pi} \left(2 + \csc \left(x\right)\right) \mathrm{dx}$

$= 6 g \cdot {\left[2 x + \ln | \left(\tan \left(\frac{x}{2}\right)\right) |\right]}_{\frac{1}{8} \pi}^{\frac{3}{4} \pi}$

=6g(3/2pi+ln(tan(3/8pi)))-(1/4pi+3ln(tan(1/16pi))#

$= 6 g \left(6.42\right)$

$= 377.5 J$