Two blocks are connected by a lightweight cord that passes over a frictionless, massless pulley. The bottom block has a mass of 2.00 kg and the top block has mass 0.500 kg. The coefficient of kinetic friction at each sliding surface is equal to 0.10. The two blocks are release from rest. Determine the acceleration of each block.

To determine the acceleration of each block, we can apply Newton's second law of motion:

F_net = m * a

where F_net is the net force acting on the block, m is the mass of the block, and a is the acceleration.

Let's start by analyzing the forces acting on each block:

For the bottom block:
- There is the force of gravity pulling it downward (F_gravity_bottom = m_bottom * g, where g is the acceleration due to gravity)
- There is the friction force opposing its motion (F_friction_bottom = μ * F_normal_bottom, where F_normal_bottom is the normal force acting on the bottom block and μ is the coefficient of kinetic friction)

For the top block:
- There is the force of gravity pulling it downward (F_gravity_top = m_top * g, where m_top is the mass of the top block)
- There is the tension force from the cord pulling it upward (F_tension)

Next, we need to relate the tension force and the friction force:

Since the cord connecting the two blocks is light, the tension force is the same throughout the cord. Therefore, the tension force pulling up on the top block is also the tension force pulling down on the bottom block. This means F_tension = Tension = F_tension_bottom = F_tension_top.

Now, let's find the net force for each block:

For the bottom block:
F_net_bottom = F_gravity_bottom - F_friction_bottom
= m_bottom * g - μ * F_normal_bottom
= m_bottom * g - μ * m_bottom * g (since F_normal_bottom = m_bottom * g)
= (m_bottom - μ * m_bottom) * g

For the top block:
F_net_top = F_gravity_top + F_tension
= m_top * g + F_tension

Since the two blocks are connected, their accelerations are the same, let's call it a. Therefore, the net force for both blocks is given by:

F_net = (m_bottom - μ * m_bottom) * g = (m_top + m_bottom) * a

Now we can solve for the acceleration:

(m_bottom - μ * m_bottom) * g = (m_top + m_bottom) * a

Substituting the given values: m_bottom = 2.00 kg, m_top = 0.500 kg, μ = 0.10, and g = 9.8 m/s^2:

(2.00 kg - 0.10 * 2.00 kg) * 9.8 m/s^2 = (0.500 kg + 2.00 kg) * a

(2.00 kg - 0.20 kg) * 9.8 m/s^2 = 2.50 kg * a

(1.80 kg) * 9.8 m/s^2 = 2.50 kg * a

17.64 kg·m/s^2 = 2.50 kg * a

Dividing both sides by 2.50 kg:

7.056 m/s^2 = a

Therefore, the acceleration of each block is 7.056 m/s^2.