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

Mar 14, 2018

The work is $= 372.8 J$

#### Explanation:

$\text{Reminder : }$

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

The work done is

$W = F \cdot d$

The frictional force is

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

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

The normal force is $N = m g$

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

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

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

The work done is

$W = 4 g {\int}_{- \frac{3}{12} \pi}^{\frac{1}{3} \pi} \left(5 + \tan x\right) \mathrm{dx}$

=4g*[5x-ln(cos(x)]_(-3/12pi)^(1/3pi)

=4g((5/3pi-ln(cos(1/3pi))-(-15/12pi-ln(cos(-3/12pi))))#

$= 4 g \left(\frac{35}{12} \pi + \frac{1}{12} \ln 64\right)$

$= 4 \cdot 9.8 \cdot 9.51$

$= 372.8 J$