An object with a mass of 6 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 1+cscx . How much work would it take to move the object over x in [(3pi)/4, (7pi)/8], where x is in meters?

May 1, 2018

The work is $= 66.21 J$

Explanation:

$\text{Reminder : }$

$\int \csc x \mathrm{dx} = \ln | \left(\tan \left(\frac{x}{2}\right)\right) | + C$

The work done is

$W = F \cdot d$

The frictional force is

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

The coefficient of kinetic friction is ${\mu}_{k} = \left(1 + \csc \left(x\right)\right)$

The normal force is $N = m g$

The mass of the object is $m = 6 k g$

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

$= 6 \cdot \left(1 + \csc \left(x\right)\right) g$

The work done is

$W = 6 g {\int}_{\frac{3}{4} \pi}^{\frac{7}{8} \pi} \left(1 + \csc \left(x\right)\right) \mathrm{dx}$

$= 6 g \cdot {\left[x + \ln | \left(\tan \left(\frac{x}{2}\right)\right) |\right]}_{\frac{3}{4} \pi}^{\frac{7}{8} \pi}$

=6g(7/8pi+ln(tan(7/16pi)))-(3/4pi+ln(tan(3/8pi))#

$= 6 g \left(1.126\right)$

$= 66.21 J$