# An object with a mass of 4 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?

Jan 24, 2018

The work is $= 61.5 J$

#### 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 ${\mu}_{k} = \left(1 + 3 \csc \left(x\right)\right)$

The normal force is $N = m g$

The mass of the object is $m = 4 k g$

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

$= 4 \cdot \left(1 + 3 \csc \left(x\right)\right) g$

The work done is

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

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

=4g(1/4pi+ln(tan(1/8pi)))-(1/6pi+ln(tan(1/12pi))#

$= 4 g \left(1.57\right)$

$= 61.5 J$