Question

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?

Answer 1

`W = 1236` `"J"`

Explanation:

We're asked to find the necessary work that needs to be done on a `3`-`"kg"` object to move on the position interval `x in [1color(white)(l)"m", 3color(white)(l)"m"]` 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

`ulbar(|stackrel(" ")(" "W = int_(x_1)^(x_2)F_xdx" ")|)` `color(white)(a)` (one dimension)

where

• `F_x` is the magnitude of the necessary force

• `x_1` is the original position (`1` `"m"`)

• `x_2` is the final position (`3` `"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_kn`

Since the surface is horizontal, `n = mg`, so

`ul(F_x = mu_kmg`

The quantity `mg` 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` `"m"` to `x=3` `"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"" ")|)`

Answer 2

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^@`

Work Done against force `vecF` for moving a distance `vecs` is given as

`W=vecFcdotvecs`

Let the object move a distanace `vecdx`. Work done to move the object against the force of friction

`dW=vec(F_f)cdot vecdx`

Force of friction `vec(F_f)=-u_kNhatx`
where `N` is normal reaction and given `=mg`
`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(54-12)`
`W=-1236J`

`-ve` sign shows that this amount of energy was lost in doing work against friction.