Question

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

Answer 1

Force exerted on the right cube by the middle cube:

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

Step-by-step 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

Forces (x) in total mass : Newton's second law

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

`a=(36-6)/(18) = 1.67 (m)/(s)`

Forces (y) in total mass : Newton's first law

∑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

Calculation of the coefficient of kinetic friction

μk=Ff/Nt Ff: friction force = 6 N

μk=6/176.4 = 0.034

Forces (x) on the first block (on the right): Newton's second law

∑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