Block A of mass 115 kg rests on a table with which it has a coefficient of limiting friction of 0.37. A light cord passes horizontally from Block A, without slipping, over a pullet of mass 34kg, radius 120 mm and radius of gyration 84mm, and down vertically of block B of mass 142 kg, which hangs freely. The system released.

The acceleration of the blocks?

The cord tension adjacent to block A?

To find the acceleration of the blocks and the cord tension adjacent to block A, we need to consider the forces acting on each block and apply Newton's second law of motion.

First, let's analyze the forces acting on Block A:
- Weight (mg): The weight of Block A is given by mass (m) multiplied by the acceleration due to gravity (g). So, the weight of Block A is (115 kg) * (9.8 m/s²) = 1127 N.
- Normal force (N): The normal force exerted by the table on Block A counteracts the weight and is equal in magnitude but opposite in direction. Therefore, the normal force is 1127 N.
- Frictional force (f): The frictional force opposing the motion is given by the coefficient of limiting friction (µ) multiplied by the normal force (N). So, the frictional force is (0.37) * (1127 N) = 416.99 N. Since this is the maximum static frictional force, it will be used to calculate the maximum force of static friction.

The force of static friction acts to prevent Block A from sliding. Therefore, the maximum static frictional force must be equal to or less than the force applied on Block A by the hanging mass.

Next, let's analyze the forces acting on Block B:
- Weight (mg): The weight of Block B is given by mass (m) multiplied by the acceleration due to gravity (g). So, the weight of Block B is (142 kg) * (9.8 m/s²) = 1392.6 N.

Considering the above information, we can now calculate the acceleration of the blocks and the cord tension adjacent to Block A.

Since the system is released, the force of static friction will change from static friction to kinetic friction. The maximum static frictional force (416.99 N) will change to the force of kinetic friction, which is given by the coefficient of kinetic friction (µk) multiplied by the normal force (N).

We can calculate the force of kinetic friction as:
Force of kinetic friction (fk) = µk * N

The relationship between the force of kinetic friction and the mass of Block B is given by:
fk = (mass of Block B) * (acceleration)

Using these equations, we can solve for the acceleration of the blocks and the cord tension adjacent to Block A.

To find the acceleration of the blocks, we need to analyze the forces acting on the system.

1. Forces acting on block A:
- Weight (mg) acting vertically downwards.
- Normal force (N) acting vertically upwards from the table.
- Frictional force (Ff) acting horizontally opposite to the direction of motion.

2. Forces acting on the pulley:
- Tension force (T) acting horizontally.
- Weight (mg) acting vertically downwards.

3. Forces acting on block B:
- Tension force (T) acting vertically upwards.
- Weight (mg) acting vertically downwards.

Considering the system as a whole, we can write the following equations:

1. For block A:
∑Fy = N - mg = 0 (since block A is in equilibrium vertically)
∑Fx = T - Ff = ma (since block A is accelerating horizontally)
Ff = μN (where μ is the coefficient of friction, and N is the normal force)

Substitute the value of N in terms of mg:
μmg = ma

2. For the pulley:
∑Fy = T - mg = 0 (since the pulley is in equilibrium vertically)

Therefore, T = mg

3. For block B:
∑Fy = T - mg = ma (since block B is accelerating vertically)

We can substitute T = mg into the equation for block B:
mg - mg = ma
a = 0

From the equations, we find that the acceleration of the blocks is 0. This means that the system is in equilibrium, and there is no net acceleration.

Now, to find the tension in the cord adjacent to block A (T), we substitute T = mg into the equation for block A:
μmg = ma

Substituting the given values:
0.37 * 115 * 9.8 = 115 * a

Simplifying:
a = 0.37 * 9.8

The acceleration of the blocks is approximately 3.626 m/s².

The tension in the cord adjacent to block A is equal to the weight of block A:
T = m * g
T = 115 kg * 9.8 m/s²

Calculating:
T ≈ 1127 N