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

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

don't remove my answer im trying my best to answer them

explanation:

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.