# An object with a mass of 7 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 4+secx . How much work would it take to move the object over x in [(pi)/12, (pi)/6], where x is in meters?

Aug 31, 2017

The work is $= 19.5 J$

#### Explanation:

We need

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

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 $m = 7 k g$

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

$= 7 \cdot \left(4 + \sec x\right) g$

The work done is

$W = 7 g {\int}_{\frac{1}{12} \pi}^{\frac{1}{6} \pi} \left(4 + \sec x\right) \mathrm{dx}$

$= 7 g \cdot {\left[\ln \left(\tan x + \sec x\right)\right]}_{\frac{1}{12} \pi}^{\frac{1}{6} \pi}$

=7g(ln(tan(pi/6)+sec(pi/6))-ln(tan(pi/12)+sec(pi/12))#

$= 7 g \left(0.284\right)$

$= 19.5 J$