An object with a mass of 5 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= x+xsinx . How much work would it take to move the object over x in [0, 8pi], where x is in meters?

Aug 10, 2017

$W \approx 14259 J$

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:

• $\mapsto \text{m"=5"kg}$
• $\mapsto {\mu}_{k} \left(x\right) = x + x \sin \left(x\right)$
• $\mapsto x \in \left[0 , 8 \pi\right]$
• $\mapsto g = 9.81 {\text{m"//"s}}^{2}$

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:

$\implies W = \left(5\right) \left(9.81\right) {\int}_{0}^{8 \pi} \left(x + x \sin x\right) \mathrm{dx}$

Let's break the integral up using the sum rule.

$\implies = \left(5\right) \left(9.81\right) \left[{\int}_{0}^{8 \pi} \left(x\right) \mathrm{dx} + {\int}_{0}^{8 \pi} \left(x \sin x\right) \mathrm{dx}\right]$

The left is a basic integral, but the right will require integration by parts to solve.

$\text{ "u=x " } \mathrm{du} = \mathrm{dx}$

$\mathrm{dv} = \sin \left(x\right) \mathrm{dx} \text{ } v = - \cos \left(x\right)$

We will have an integral of the form:

$u v - v \int \mathrm{du}$

Therefore, for the right most integral, we obtain:

$- x \cos \left(x\right) - \left({\int}_{0}^{8 \pi} \left(- \cos x\right) \mathrm{dx}\right)$

$\implies - x \cos \left(x\right) + {\int}_{0}^{8 \pi} \left(\cos x\right) \mathrm{dx}$

We now have a basic integral.

Putting it all together:

$\left(5\right) \left(9.81\right) \left[{\int}_{0}^{8 \pi} \left(x\right) \mathrm{dx} + \left\{- x \cos \left(x\right) + {\int}_{0}^{8 \pi} \left(\cos x\right) \mathrm{dx}\right\}\right]$

Which we can now evaluate, yielding $W = 14258.57$

Therefore, we have that the work done is $\approx 14259 J$ or $1.4 \cdot {10}^{4} J$