Let's solve this problem step by step.

We are given:

• Mass of block X, $m_X = 2.0 \, \text{kg}$
• Mass of block Y, $m_Y = 6.0 \, \text{kg}$
• A force of $F = 24 \, \text{N}$ acts on block X.
• The blocks are in contact and on a frictionless surface.

1. Determine the acceleration of the system:

Since the two blocks are in contact and the surface is frictionless, they will accelerate together as a single system. The total mass of the system is:

$m_{\text{total}} = m_X + m_Y = 2.0 \, \text{kg} + 6.0 \, \text{kg} = 8.0 \, \text{kg}$

Using Newton's second law, the acceleration $a$ of the system is:

$a = \frac{F}{m_{\text{total}}} = \frac{24 \, \text{N}}{8.0 \, \text{kg}} = 3.0 \, \text{m/s}^2$

2. Determine the force exerted by block X on block Y:

Now that we have the acceleration, we can find the force that block X exerts on block Y.

Since both blocks accelerate at the same rate, the force exerted by block X on block Y must be equal to the force required to accelerate block Y at the same acceleration.

Using Newton's second law for block Y:

$F_{\text{XY}} = m_Y \times a = 6.0 \, \text{kg} \times 3.0 \, \text{m/s}^2 = 18.0 \, \text{N}$

So, the force that block X exerts on block Y is 18 N.

Would you like further details on any part of this solution or have more questions?

Here are 5 related questions you can explore:

1. What is the reaction force that block Y exerts on block X?
2. How would the acceleration change if block Y's mass was doubled?
3. What would happen if the surface was not frictionless? How would friction affect the forces?
4. How does Newton's third law apply to the interaction between block X and block Y?
5. How would the force diagram look for each block individually?

Tip: Always double-check the net force and acceleration in systems involving multiple objects. Treat the entire system as one unit before analyzing individual interactions.