# 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 6, 2017

The work is $= 257 J$

#### Explanation:

The work done is

$W = F \cdot d$

The frictional force is

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

The normal force is $N = m g$

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

=4(5+tanx))g

The work done is

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

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

$= 4 g \left(\left(\frac{25}{12} \pi - \ln | \cos \left(\frac{5}{12} \pi\right) |\right) - \left(\frac{5}{12} \pi - \ln | \cos \left(\frac{1}{12} \pi\right) |\right)\right)$

=4g(5/3pi-ln|cos(5/12pi)|+ln(cos(1/12pi))

=4g(5/3pi+1.32))#

$= 4 g \cdot 6.56 J$

$= 257 J$