# 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?

Aug 7, 2017

${a}_{x} = 1.66$ ${\text{m/s}}^{2}$

#### Explanation:

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

$m = 65$ $\text{kg}$ $+ 125$ $\text{kg}$ = ul(190color(white)(l)"kg"

There are two horizontal forces acting on the crate:

the applied force (${F}_{\text{applied}}$) 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

$\sum {F}_{x} = {F}_{\text{applied}} - {f}_{k} = m {a}_{x}$

The friction force is given by the equation

${f}_{k} = {\mu}_{k} n$

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, $m g$:

${f}_{k} = {\mu}_{k} m g$

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

• ${F}_{\text{applied}} = 650$ $\text{N}$

• ${\mu}_{k} = 0.18$

• $m = 190$ $\text{kg}$

• and $g = 9.81$ ${\text{m/s}}^{2}$

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" ")|)