Question

An object with a mass of `5 kg` is pushed along a linear path with a kinetic friction coefficient of `u_k(x)= e^x-2x+3`. How much work would it take to move the object over #x in [3, 4], where x is in meters?

Answer 1

`W = 1497` `"J"`

Explanation:

We're asked to find the necessary work that needs to be done on a `5`-`"kg"` object to move on the position interval `x in [3color(white)(l)"m", 4color(white)(l)"m"]` with a varying coefficient of kinetic friction `mu_k` represented by the equation

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

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 (`3` `"m"`)

• `x_2` is the final position (`4` `"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 = (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")(e^x-2x+3)`

Work is the integral of force, and we're measuring it from `x=3` `"m"` to `x=4` `"m"`, so we can write

`color(red)(W) = int_(3color(white)(l)"m")^(4color(white)(l)"m")(49.05color(white)(l)"N")(e^x-2x+3) dx = color(red)(ulbar(|stackrel(" ")(" "1497color(white)(l)"J"" ")|)`