An object with a mass of 6 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 2+cscx . How much work would it take to move the object over x in [pi/12, (3pi)/4], where x is in meters?

Jun 4, 2017

The work is $= 417.5 J d$

Explanation:

We need

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

The work done is

$W = F \cdot d$

The frictional force is

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

The normal force is $N = m g$

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

=6(2+cscx))g

The work done is

$W = 6 g {\int}_{\frac{1}{12} \pi}^{\frac{3}{4} \pi} \left(2 + \csc x\right) \mathrm{dx}$

$= 6 g \cdot {\left[2 x + \ln | \tan \left(\frac{x}{2}\right) |\right]}_{\frac{1}{12} \pi}^{\frac{3}{4} \pi}$

=6g((3/2pi+ln|tan(3/8pi)|)-(1/6pi+ln|tan(pi/24)|)#

$= 6 g \left(\frac{4}{3} \pi + 0.88 + 2.027\right)$

$= 6 g \left(7.098\right)$

$= 42.6 g J$

$= 417.5 J$