Two objects are in contact on a frictionless surface. A horizontal force is applied to one object as shown in the figure below. Use m1 = 0.4 g, m2 = 223 g and F = 0.7 N to find the force between the two objects.

Thanks

Since there is no "figure below", you need to tell us which object is being pushed by F.

Assume both objects accelerate at the same rate. The net force on each object must then be proportional to its mass. You can use that fact to compute the force between them

To find the force between the two objects, we can use Newton's second law of motion, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration. In this case, since the objects are on a frictionless surface, the acceleration of the objects will be the same.

Let's assign m1 to be the mass of the first object and m2 to be the mass of the second object. We are given that m1 = 0.4 g and m2 = 223 g. To use Newton's second law, we need to convert these masses to kilograms since the SI unit for mass is kilograms.

Converting m1 to kilograms:
m1 = 0.4 g = 0.4 / 1000 kg (since there are 1000 grams in a kilogram)
m1 = 0.0004 kg

Converting m2 to kilograms:
m2 = 223 g = 223 / 1000 kg
m2 = 0.223 kg

Next, we can use the formula F = m * a, where F is the net force, m is the mass, and a is the acceleration.

Since the objects are in contact, they experience the same force. So, the force applied to one object is the force experienced by the other object.

We are given F = 0.7 N, so we can rearrange the formula to solve for the acceleration:

a = F / m

For both objects, the mass is the same, so we can rewrite the equation as:

a = F / (m1 + m2)

Substituting the given values:

a = 0.7 N / (0.0004 kg + 0.223 kg)
a = 0.7 N / 0.2234 kg
a ā‰ˆ 3.13 m/s^2

Now, we can find the force between the two objects by multiplying the acceleration by the mass of either object:

Force = m1 * a (or Force = m2 * a)

Force = 0.0004 kg * 3.13 m/s^2
Force ā‰ˆ 0.00125 N

Therefore, the force between the two objects is approximately 0.00125 N.