Question

How much work would it take to horizontally accelerate an object with a mass of `8 kg` to `3 m/s` on a surface with a kinetic friction coefficient of `6`?

Answer 1

`W = 36` `"J"` `+ 471color(white)(l)"N" xx "distance"`

Explanation:

We're asked to find the necessary work to accelerate an `8`-`"kg"` object from `0` to `3` `"m/s"` on a surface with a coefficient of kinetic friction of `6`.

The equation for work according to the work-energy theorem is given by

`ul(W = DeltaE_k = 1/2m(v_2)^2 - 1/2m(v_1)^2`

The final velocity is `3` `"m/s"`, and its mass is `8` `"kg"`, so we have

`W = 1/2(8color(white)(l)"kg")(3color(white)(l)"m/s" - 0)^2 = color(red)(ul(36color(white)(l)"J"`

However, this does not include the work done to counteract the friction, so we have to use the equation

`ul(W = Fs`

The necessary force `F` is

`F = f_k = mu_kmg = (6)(8color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = 471` `"N"`

And so we have

`W = (471color(white)(l)"N")s`

The distance it travels `s` can be any value because the applied force can be any number greater than the necessary force. Thus, we can leave it in terms of `s`:

`W_"necessary" = color(blue)(ulbar(|stackrel(" ")(" "36color(white)(l)"J" + (471color(white)(l)"N")s" ")|)`