An object with a mass of 3 kg is pushed along a linear path with a kinetic friction coefficient of u_k(x)= 3x^2-2x+12 . How much work would it take to move the object over x in [1, 3], where x is in meters?

Aug 5, 2017

$W = 1236$ $\text{J}$

Explanation:

We're asked to find the necessary work that needs to be done on a $3$-$\text{kg}$ object to move on the position interval $x \in \left[1 \textcolor{w h i t e}{l} \text{m", 3color(white)(l)"m}\right]$ with a varying coefficient of kinetic friction ${\mu}_{k}$ represented by the equation

ul(mu_k(x) = 3x^2-2x+12

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 ($1$ $\text{m}$)

• ${x}_{2}$ 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 = (3color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = 29.43color(white)(l)"N"

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

ul(F_x = (29.43color(white)(l)"N")(3x^2-2x+12)

Work is the integral of force, and we're measuring it from $x = 1$ $\text{m}$ to $x = 3$ $\text{m}$, so we can write

color(red)(W) = int_(1color(white)(l)"m")^(3color(white)(l)"m")(29.43color(white)(l)"N")(3x^2-2x+12) dx = color(red)(ulbar(|stackrel(" ")(" "1236color(white)(l)"J"" ")|)

Aug 5, 2017

We observe that Coefficient of kinetic friction is given as

u_k(x)=3x^2−2x+12

and that the object is being pushed along a linear path.

We know that force of friction opposes the motion. As such we can infer that movement is only in the $x$-direction.
The angle between force of friction and displacement is ${180}^{\circ}$

Work Done against force $\vec{F}$ for moving a distance $\vec{s}$ is given as

$W = \vec{F} \cdot \vec{s}$

Let the object move a distanace $\vec{\mathrm{dx}}$. Work done to move the object against the force of friction

$\mathrm{dW} = \vec{{F}_{f}} \cdot \vec{\mathrm{dx}}$

Force of friction $\vec{{F}_{f}} = - {u}_{k} N \hat{x}$
where $N$ is normal reaction and given $= m g$
dW=-(3x^2−2x+12)mghatxcdot vecdx

Integrating both sides for the given interval we get
W=-int_1^3(3x^2−2x+12)mgdx

W=-3xx9.81|3x^3/3−2x^2/2+12x|_1^3
W=-29.43[(3^3−3^2+12xx3)-(1^3−1^2+12xx1)]
W=-29.43[(27−9+36)-(1−1+12)]#
$W = - 29.43 \left(54 - 12\right)$
$W = - 1236 J$

$- v e$ sign shows that this amount of energy was lost in doing work against friction.