An object with a mass of 6 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 1+cotx . How much work would it take to move the object over x in [(pi)/12, (5pi)/8], where x is in meters?

Jul 24, 2018

Answer:

The work is $= 174.64 J$

Explanation:

$\text{Reminder : }$

$\int \cot x \mathrm{dx} = \ln | \left(\sin \left(x\right)\right) | + C$

The work done is

$W = F \cdot d$

The frictional force is

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

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

The normal force is $N = m g$

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

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

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

The work done is

$W = 6 g {\int}_{\frac{1}{12} \pi}^{\frac{5}{8} \pi} \left(1 + \cot \left(x\right)\right) \mathrm{dx}$

$= 6 g \cdot {\left[x + \ln | \left(\sin \left(x\right)\right) |\right]}_{\frac{1}{12} \pi}^{\frac{5}{8} \pi}$

=6g((5/8pi+ln(sin(5/8pi))-1/12pi-ln(sin(1/12pi)))#

$= 6 g \left(2.97\right)$

$= 174.64 J$