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?

1 Answer
Aug 7, 2017

Answer:

#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" ")|)#