An object with a mass of 6 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 1+5cotx . How much work would it take to move the object over x in [(11pi)/12, (7pi)/8], where x is in meters?

Aug 17, 2017

$W \approx 107 \text{J}$

Explanation:

Work done by a variable force is defined as the area under the force vs. displacement curve. Therefore, it can be expressed as the integral of force:

$\textcolor{\mathrm{da} r k b l u e}{W = {\int}_{{x}_{i}}^{{x}_{f}} {F}_{x} \mathrm{dx}}$

where ${x}_{i}$ is the object's initial position and ${x}_{f}$ is the object's final position

Assuming dynamic equilibrium, where we are applying enough force to move the object but not enough to cause a net force and therefore not enough to cause an acceleration, our parallel forces can be summed as:

$\sum {F}_{x} = {F}_{a} - {f}_{k} = 0$

Therefore we have that ${F}_{a} = {f}_{k}$

We also have a state of dynamic equilibrium between our perpendicular forces:

$\sum {F}_{y} = n - {F}_{g} = 0$

$\implies n = m g$

We know that ${\vec{f}}_{k} = {\mu}_{k} \vec{n}$, so putting it all together, we have:

${\vec{f}}_{k} = {\mu}_{k} m g$

$\implies \textcolor{\mathrm{da} r k b l u e}{W = {\int}_{{x}_{i}}^{{x}_{f}} {\mu}_{k} m g \mathrm{dx}}$

We have the following information:

• $\mapsto \text{m"=6"kg}$
• $\mapsto {\mu}_{k} \left(x\right) = 1 + 5 \cot \left(x\right)$
• $\mapsto x \in \left[\frac{11 \pi}{12} , \frac{7 \pi}{8}\right]$
• $\mapsto g = 9.81 {\text{m"//"s}}^{2}$

Returning to our integration, know the $m g$ quantity, which we can treat as a constant and move outside the integral.

$\textcolor{\mathrm{da} r k b l u e}{W = m g {\int}_{{x}_{i}}^{{x}_{f}} {\mu}_{k} \mathrm{dx}}$

Substituting in our known values:

$\implies W = \left(6\right) \left(9.81\right) {\int}_{\frac{11 \pi}{12}}^{\frac{7 \pi}{8}} 1 + 5 \cot \left(x\right) \mathrm{dx}$

This is a basic integral.

$\implies W \approx 107.39$

Therefore, we have that the work done is $\approx 107 J$.