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

Aug 5, 2017

$W = 335$ $\text{J}$

#### Explanation:

We're asked to find the necessary work that needs to be done on a $5$-$\text{kg}$ object to move on the position interval $x \in \left[2 \textcolor{w h i t e}{l} \text{m", 3color(white)(l)"m}\right]$ with a varying coefficient of kinetic friction ${\mu}_{k}$ represented by the equation

ul(mu_k(x) = x^2-x+3

The work $W$ done by the necessary force is given by

$\underline{\overline{| \stackrel{\text{ ")(" "W = int_(x_1)^(x_2)F_xdx" }}{|}}}$ $\textcolor{w h i t e}{a}$ (one dimension)

where

• ${F}_{x}$ is the magnitude of the necessary force

• ${x}_{1}$ is the original position ($2$ $\text{m}$)

• ${x}_{2}$ is the final position ($3$ $\text{m}$)

The necessary force would need to be equal to (or greater than, but we're looking for the minimum value) the retarding friction force, so

${F}_{x} = {f}_{k} = {\mu}_{k} n$

Since the surface is horizontal, $n = m g$, so

ul(F_x = mu_kmg

The quantity $m g$ equals

mg = (5color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = 49.05color(white)(l)"N"

And we also plug in the above coefficient of kinetic friction equation:

ul(F_x = (49.05color(white)(l)"N")(x^2-x+3)

Work is the integral of force, and we're measuring it from $x = 3$ $\text{m}$ to $x = 4$ $\text{m}$, so we can write

color(red)(W) = int_(2color(white)(l)"m")^(3color(white)(l)"m")(49.05color(white)(l)"N")(x^2-x+3) dx = color(red)(ulbar(|stackrel(" ")(" "335color(white)(l)"J"" ")|)#