An object with a mass of 8 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 5x^2-x+1 . How much work would it take to move the object over #x in [2, 3], where x is in meters?

Aug 6, 2017

$W \approx 2367 J$

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:

$W = {\int}_{{x}_{i}}^{{x}_{f}} {F}_{x} \mathrm{dx}$

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

$\implies W = {\int}_{{x}_{i}}^{{x}_{f}} {\mu}_{k} m g \mathrm{dx}$

We know the $m g$ quantity, which we can treat as a constant and move outside the integral.

$\implies W = \left(8\right) \left(9.81\right) {\int}_{2}^{3} \left(5 {x}^{2} - x + 1\right) \mathrm{dx}$

This is a basic integral, yielding $W = 2367.48$

Therefore, we have that the work done is $\approx 2367 J$ or $2.4 \cdot {10}^{3} J$