A 4kg wooden block is placed on a wooden table, so the coeficent of friction would be 0.2. There is a string connected to both sides of the wooden block with a pulley system at the end of the table to go off and then each string has a different weight attached to it. On the left side there is a block with a mass of 3kg attached to it and on the right side there is a block with a mass of 5kg attached to it. What is the acceleration of the system and the tension in each string?

To find the acceleration of the system and the tension in each string, we can use Newton's laws of motion.

First, let's analyze the forces acting on the system.

1. Weight of the 4kg wooden block:
The weight acts vertically downwards with a force equal to the mass of the block (4kg) multiplied by the acceleration due to gravity (9.8 m/s^2). So, the weight of the block is 4 kg x 9.8 m/s^2 = 39.2 N.

2. Normal force:
The normal force acts perpendicular to the table, and its magnitude is equal to the weight of the block (since the block is not accelerating vertically). Therefore, the normal force is also 39.2 N.

3. Frictional force:
The frictional force opposes the motion of the block and is given by the equation: Frictional force = coefficient of friction x normal force.
Thus, the frictional force is 0.2 x 39.2 N = 7.84 N.

Now, let's consider each side of the pulley system separately.

Left side:
The forces acting on the 3kg block are its weight (3 kg x 9.8 m/s^2 = 29.4 N) and the tension in the string (T1). The net force on the 3kg block is given by:
Net force = T1 - 29.4 N.

Right side:
The forces acting on the 5kg block are its weight (5 kg x 9.8 m/s^2 = 49 N), the tension in the string (T2), and the frictional force (7.84 N). The net force on the 5kg block is given by:
Net force = T2 - 49 N - 7.84 N.

Since the two blocks are connected by the same string, the tension in the string on both sides of the pulley system is equal:
T1 = T2.

Now, let's calculate the acceleration of the system using Newton's second law, which states that the net force acting on the system is equal to the mass of the system multiplied by its acceleration.

The net force acting on the system is the sum of the net forces on each block:
Net force system = T1 - 29.4 N + T2 - 49 N - 7.84 N.

Since the acceleration of the system is the same for both blocks, we can write:
Net force system = total mass x acceleration.

The total mass of the system is the sum of the masses of the two blocks and the 4kg wooden block:
Total mass = 3 kg + 5 kg + 4 kg = 12 kg.

Putting all the values together, we have:
T1 - 29.4 N + T2 - 49 N - 7.84 N = 12 kg x acceleration.

However, we also know that T1 = T2. So, we can rewrite the equation as:
2T1 - 76.24 N = 12 kg x acceleration.

To find the tension in each string (T1 and T2), we need another equation. Let's consider the torque around the pulley.

The torque exerted by the 3kg block (clockwise) is given by:
Torque3kg = T1 x (radius of pulley) = T1 x r.

The torque exerted by the 5kg block (counterclockwise) is given by:
Torque5kg = T2 x (radius of pulley) = T2 x r.

Since the pulley is not accelerating, the sum of the torques must be zero:
Torque3kg - Torque5kg = 0.

Plugging in the values, we have:
T1 x r - T2 x r = 0.

Simplifying this equation, we find:
T1 = T2.

Now, we can substitute T1 for T2 in the net force equation:
2T1 - 76.24 N = 12 kg x acceleration.

Combining the equations, we have:
2T1 - 76.24 N = 12 kg x acceleration.

Now, we can solve for the acceleration and tension:
2T1 = 12 kg x acceleration + 76.24 N
T1 = (12 kg x acceleration + 76.24 N) / 2

To summarize:
The acceleration of the system is given by (12 kg x acceleration + 76.24 N) / 2.
The tension in each string, T1 and T2, is the same and can be found using T1 = (12 kg x acceleration + 76.24 N) / 2.