Question

Two crates, of mass 65 kg and 125 kg, are in contact and at rest on a horizontal surface, A 650-N force is exerted on the 65 kg crate, the coefficient of kinetic friction is 0.18. What is the acceleration of the system?

Answer 1

`a_x = 1.66` `"m/s"^2`

Explanation:

The two objects will move as one body, so we can picture them as a single composite body with mass

`m = 65` `"kg"` `+ 125` `"kg"` `= ul(190color(white)(l)"kg"`

There are two horizontal forces acting on the crate:

the applied force (`F_"applied"`) directed in we'll say the positive direction

the retarding kinetic friction force (`f_k`), directed in the negative direction because it opposes motion

The net horizontal force equation is thus

`sumF_x = F_"applied" - f_k = ma_x`

The friction force is given by the equation

`f_k = mu_kn`

where

• `mu_k` is the coefficient of kinetic friction

• `n` is the magnitude of the upward normal force exerted by the surface, which since it is horizontal, is equal in magnitude to its weight, `mg`:

`f_k = mu_kmg`

Substituting this into the net force equation above:

`ul(sumF_x = F_"applied" - mu_kmg = ma_x`

Now, let's solve for the acceleration, `a_x`:

`color(red)(a_x = (F_"applied" - mu_kmg)/m`

The problem gives us

• `F_"applied" = 650` `"N"`

• `mu_k = 0.18`

• `m = 190` `"kg"`

• and `g = 9.81` `"m/s"^2`

Plugging these in:

`a_x = (650color(white)(l)"N" - 0.18(190color(white)(l)"kg")(9.81color(white)(l)"m/s"^2))/(190color(white)(l)"kg") = color(blue)(ulbar(|stackrel(" ")(" "1.66color(white)(l)"m/s"^2" ")|)`