# An object with a mass of 4 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 5+tanx . How much work would it take to move the object over #x in [(-5pi)/12, (5pi)/12], where x is in meters?

Aug 12, 2017

The work is $= 513.1 J$

#### Explanation:

We need

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

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 $m = 4 k g$

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

$= 4 \left(5 + \tan x\right) g$

The work done is

$W = 4 g {\int}_{- \frac{5}{12} \pi}^{\frac{5}{12} \pi} \left(5 + \tan x\right) \mathrm{dx}$

$= 4 g \cdot {\left[5 x + \ln | \cos \left(x\right) |\right]}_{- \frac{5}{12} \pi}^{\frac{5}{12} \pi}$

$= 4 g \left(\left(5 \cdot \frac{5}{12} \pi + \ln | \cos \left(\frac{5}{12} \pi\right) |\right) - \left(5 \cdot - \frac{5}{12} \pi + \ln | \cos \left(- \frac{5}{12} \pi\right) |\right)\right)$

$= 4 g \left(\frac{50}{12} \pi + \ln | \cos \left(\frac{5}{12} \pi\right) | - \ln | \cos \left(\frac{5}{12} \pi\right) |\right)$

$= 4 g \left(\frac{25}{6} \pi + 0\right)$

$= 513.1 J$