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?

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?

Isabella Alvarez · Accepted Answer

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

We can think of the two objects as a single composite body with mass because they will move as one body.

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

The crate is being acted upon by two horizontal forces:

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

Therefore, the equation for net horizontal force is

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

Equation provides the friction force.

#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#

Adding this to the previous net force equation:

#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 issue provides us with

#F_"applied" = 650# #"N"#

#mu_k = 0.18#

#m = 190# #"kg"#

and #g = 9.81# #"m/s"^2#

Connecting these:

#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" ")|)#

Benjamin Asiedu · Answer

undefined To find the acceleration of the system, first calculate the net force acting on the system. Then, use Newton's second law (F = ma) to find the acceleration.

1. Calculate the force of friction acting on the 65 kg crate:
   $ F_{\text{friction}} = \mu_k \times F_{\text{normal}} $
   $ F_{\text{friction}} = 0.18 \times (65 \times 9.8) $
   $ F_{\text{friction}} = 113.4 \, \text{N} $

2. Calculate the net force on the system:
   $ F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} $
   $ F_{\text{net}} = 650 \, \text{N} - 113.4 \, \text{N} $
   $ F_{\text{net}} = 536.6 \, \text{N} $

3. Calculate the total mass of the system:
   $ m_{\text{total}} = m_1 + m_2 $
   $ m_{\text{total}} = 65 \, \text{kg} + 125 \, \text{kg} $
   $ m_{\text{total}} = 190 \, \text{kg} $

4. Use Newton's second law to find the acceleration:
   $ a = \frac{F_{\text{net}}}{m_{\text{total}}} $
   $ a = \frac{536.6 \, \text{N}}{190 \, \text{kg}} $
   $ a ≈ 2.82 \, \text{m/s}^2 $

The acceleration of the system is approximately $ 2.82 \, \text{m/s}^2 $.