The horizontal surface on which the objects slide is frictionless. If F = 6.0 N and M = 1.0 kg, what is the magnitude of the force exerted on the large block by the small block?

The image is missing, so i have attached it.

The force exerted on the large block by the small block = 8.4 N

Explanation:

From the image attached, the mass of the small block = 2M while the mass of the large block = 3M

Also,Force on small block = F and force on large block = 2F

Equilibrium of forces on the left gives;

2F - N = 3Ma

Thus,

Ma = (2F - N)/3 - - - - eq1

Also, on right hand side, Equilibrium of forces gives;

N - F = 2Ma

Ma = (N - F)/2 - - - - eq2

Equating eq(1) and eq(2) gives us;

(2F - N)/3 = (N - F)/2

Where N is the force exerted on the large block by the small block.

Making N the subject gives;

4F - 2N = 3N - 3F

5N = 7F

N = 7F/5

We are given F = 6N

Thus;

N = 7(6)/5

N = 8.4 N

which three statements are true about all alloys?