# 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, (pi)/4], where x is in meters?

Nov 1, 2017

The work is $= 217.6 J$

#### Explanation:

We need

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

The work done is

$W = F \cdot d$

The frictional force is

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

The normal force is $N = m g$

The mass is $m = 4 k g$

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

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

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

The work done is

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

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

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

=4g(5.55))

$= 217.6 J$