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 force is applied to the left side of the left cube causing all three cubes to accelerate to the right at 2.0 m/s2 . What is the magnitude of the force exerted on the middle cube by the left cube in this case?

Respuesta :

Answer:

24 N

Explanation:

[tex]m[/tex] = mass of the cube = [tex]6.0 kg[/tex]

Consider the three cubes together as one.

[tex]M[/tex] = mass of the three cubes together = [tex]3 m = 3 (6.0) = 18 kg [/tex]

[tex]a[/tex] = acceleration of the combination = 2 ms⁻²

[tex]F[/tex] = Force applied on the combination

Using Newton's second law

[tex]F = ma = (18) (2) = 36 N[/tex]

[tex]F_{L}[/tex] = Force by the left cube on the middle cube

Consider the forces acting on left cube, from the force diagram, we have

[tex]F - F_{L} = ma \\36 - F_{L} = (6) (2)\\F_{L} = 24 N[/tex]

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