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

Jan 2, 2018

The work done is $= 588.6 J$

Explanation:

$\text{Reminder : }$

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

The work done is

$W = F \cdot d$

The frictional force is

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

The coefficient of kinetic friction is ${\mu}_{k} = \left(4 + \sec x\right)$

The normal force is $N = m g$

The mass of the object is $m = 7 k g$

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

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

The work done is

$W = 7 g {\int}_{- \frac{5}{12} \pi}^{\frac{1}{12} \pi} \left(4 + \sec x\right) \mathrm{dx}$

$= 7 g \cdot {\left[4 x + \ln \left(\tan x + \sec x\right)\right]}_{- \frac{5}{12} \pi}^{\frac{1}{12} \pi}$

=7g(1/3pi+ln(tan(pi/12)+sec(pi/12))-(-5/3pi+ln(tan(-5/12pi)+sec(-5/12pi))#

$= 7 g \left(8.58\right)$

$= 588.6 J$
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