Two crates, of mass 65 kg and 125 kg, are in contact and at rest on a horizontal surface, A 650N 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
Answer:
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 directionthe 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,
#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" "))#