An object with a mass of 4 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 5+tanx . How much work would it take to move the object over x in [(pi)/12, (pi)/4], where x is in meters?

Aug 17, 2017

$W \approx 115 \text{J}$

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:

• $\mapsto \text{m"=4"kg}$
• $\mapsto {\mu}_{k} \left(x\right) = 5 + \tan \left(x\right)$
• $\mapsto x \in \left[\frac{\pi}{12} , \frac{\pi}{4}\right]$
• $\mapsto g = 9.81 {\text{m"//"s}}^{2}$

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:

$\implies W = \left(4\right) \left(9.81\right) {\int}_{\frac{\pi}{12}}^{\frac{\pi}{4}} 5 + \tan \left(x\right) \mathrm{dx}$

Evaluating:

$\implies \left(4\right) \left(9.81\right) {\left[5 x + \ln \left\mid \sec \left(x\right) \right\mid\right]}_{\frac{\pi}{12}}^{\frac{\pi}{4}}$

=>(39.24)[((5pi)/4+lnabs(sqrt2))-((5pi)/12+lnabs(sec(pi/12))]

=>(39.24)[(5pi)/6+lnabs(sqrt2)-lnabs(sec(pi/12)))]

$\implies \approx 114.93$

Therefore, we have that the work done is $\approx 115 J$.