Question

How much work would it take to horizontally accelerate an object with a mass of `8 kg` to `7 m/s` on a surface with a kinetic friction coefficient of `9`?

Answer 1

`W = 196` `"J"`

Explanation:

We're asked to find the total work done on an object so that it accelerates to `7` `"m/s"`, with a coefficient of kinetic friction of `9`.

Using the equation

`f_k = mu_kn`,

we can find the retarding friction force the surface exerts on the object (knowing that the normal force `n = mg`):

`f_k = mu_kmg = (9)(8color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = color(red)(706.32` `color(red)("N"`

To make this simple, we'll make the net force experienced by the object `1` `"N"`, so the applied force will be `707.32` `"N"`:

`sumF_x = 707.32` `"N"` `- 706.32` `"N"` `= 1` `"N"`

Now, we can use Newton's second law to find the acceleration of the object:

`sumF_x = ma_x`

`a_x = (sumF_x)/m = (1color(white)(l)"N")/(8color(white)(l)"kg") = 0.125` `"m/s"^2`

Now, let's use a kinematics equation to find the distance traveled under this constant acceleration so that the final velocity is `7` `"m/s"`:

`(v_x)^2 = (v_(0x))^2 + 2a_x(Deltax)`

Starting from an assumed state of rest:

`(7color(white)(l)"m/s")^2 = 0 + 2(0.125color(white)(l)"m/s"^2)(Deltax)`

`Deltax = 196` `"m"`

Finally, using the equation

`W = Fs` (one dimensional)

(`F` is the net force), we have

`W = (1color(white)(l)"N")(196color(white)(l)"m") = color(blue)(ul(196color(white)(l)"J"`