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. if the coefficient of kinetic friction is 0.18, calculate the (a) the acceleration of the system, and (b) the force that each crate exert on the other. (c) repeat with the crates reversed.

1.65 m/s^2 is the system's acceleration. 426.75 N is the force that one crate applies to the other. When the containers are turned around, a force of 221.91 N is applied by one crate to the other.In mechanics, acceleration is the rate at which an object's velocity with respect to time changes. The vector quantity of acceleration (in that they have magnitude and direction). The direction of the net force imposed on an object determines its acceleration in relation to that force.It may sound difficult, but velocity is simply the act of moving quickly in a particular direction. A velocity of an item is defined as the rate of change of the object's location with regard to a frame of reference and time.Given that,Mass of first crates = 65 kgMass of second crates = 125 kgForce = 650 NCoefficient of kinetic friction = 0.18a = 650/(65+120) - 0.18*9.8a= 1.65 m/s²The force that each crate exerts on the otherF= 125*1.65*0.18*125*9.8F= 426.75NThe force that each crate exerts on the other when the crates reversedF=125*1.65*0.18*125*9.8F=221.91NLearn more about acceleration herehttps://brainly.com/question/3046924#SPJ4