Problem:
Two masses, 2kg and 3kg, are attached to the end of a string passed over a pulley fixed at the top. Determine the tension in the string and the acceleration of the system.

Solution:

Given data:
Mass 1 (m1) = 2kg
Mass 2 (m2) = 3kg

Assumptions:
1. The pulley is massless and frictionless.
2. The string is massless and inextensible.
3. The acceleration due to gravity is 9.8 m/s^2.

Analysis:
To find the tension in the string and the acceleration of the system, we can apply Newton's second law of motion for both masses.

For mass 1 (m1):
The net force acting on mass 1 is the tension in the string (T) directed upwards, and the force due to its weight (mg) directed downwards.
Therefore, the equation of motion for mass 1 is:
T - m1g = m1a

For mass 2 (m2):
The net force acting on mass 2 is the force due to its weight (m2g) directed downwards, and the tension in the string (T) directed upwards.
Therefore, the equation of motion for mass 2 is:
m2g - T = m2a

Applying Newton's Second Law:
- We can substitute the value of acceleration (a) from the first equation into the second equation to eliminate the acceleration term.
- Rearranging the equation, we get:
m2g - T = m2(T - m1g)/m1

Calculating the acceleration:
To solve the above equation, we need to know the value of g, which is approximately 9.8 m/s^2.
Let's substitute the values into the equation and solve for acceleration (a):

3(9.8) - T = 3(T - 2(9.8))/2

29.4 - T = 3T - 58.8
4T = 88.2
T = 22.05 N

Calculating the tension:
Now, let's substitute the value of T into any of the equations to find the acceleration (a).
Using the first equation:

T - m1g = m1a
22.05 - 2(9.8) = 2a
22.05 - 19.6 = 2a
2.45 = 2a
a = 1.225 m/s^2

Final Results:
The tension in the string is 22.05 N, and the acceleration of the system is 1.225 m/s^2.

Tension is 10N and acc. is 2