# 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 [(-pi)/12, (5pi)/12], where x is in meters?

Jun 25, 2018

#### Answer:

The work is $= 359.5 J$

#### Explanation:

$\text{Reminder : }$

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

The work done is

$W = F \cdot d$

The frictional force is

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

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

The normal force is $N = m g$

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

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

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

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

The work done is

$W = 4 g {\int}_{- \frac{1}{12} \pi}^{\frac{5}{12} \pi} \left(5 + \tan \left(x\right)\right) \mathrm{dx}$

$= 4 g \cdot {\left[5 x + \ln | \left(\cos \left(x\right)\right) |\right]}_{- \frac{1}{12} \pi}^{\frac{5}{12} \pi}$

=4g(25/12pi+ln(cos(5/12pi)))-(-5/12pi+ln(cos(-1/12pi))#

$= 4 g \left(9.17\right)$

$= 359.5 J$