Three identical 6.0-kg cubes are placed on a horizontal frictionless surface in contact with one another. the cubes are lined up from left to right and a 36-n force is applied to the left side of the left cube causing all three cubes to accelerate to the right. if the cubes are each subject to a frictional force of 6.0 n, what is the magnitude of the force exerted on the right cube by the middle cube in this case

Force exerted on the right cube by the middle cube:

F= 12.02N : in the positive direction of the x axis( +x)

Explanation:

We apply Newton's second law for forces in the direction of the x-axis.

∑Fx= m*a

∑Fx:  algebraic sum of forces ( + to the right, - to the left)

m: mass

a: acceleration

We apply Newton's first law for forces in the direction of the y-axis.

∑Fx= mt*a   , mt: total mass = 6*3= 18 kg

36-6= 18*a

∑Fy= 0

Nt-Wt=0   , Nt=Wt

Nt: total normal   ,  Wt= total Weight: mt*g  , g: acceleration due to gravity

Wt=18*9,8=176.4 N

Nt=176.4 N

μk=Ff/Nt Ff: friction force = 6 N

μk=6/176.4 = 0.034

∑Fx= m₁*a

F-Ff₁= 6*1.67 , Equation  (1)

F:Force exerted on the right cube by the middle cube

Ff₁= μk*N₁ , N₁=W₁ = 9.8*6= 58.8 N

Ff₁= 0.034*58.8 = 2 N

In the  Equation  (1):

F-2= 10.02

F= 2+10.02= 12.02N

F= 12.02N

d is the correct answer

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