Problem Statement:
Two spheres of masses 3kg and 4kg are attached to ends of a string which passes over a frictionless pulley. Take g=9.8. Find the relative acceleration of the system.

Solution:
To find the relative acceleration of the system, we need to analyze the forces acting on the spheres and apply Newton's second law of motion.

Forces Acting on the System:
1. Tension force (T) in the string: The tension force in the string acts on both spheres in the upward direction.
2. Weight force (mg) of each sphere: The weight force acts downwards due to the gravitational force.

Acceleration of Sphere 1:
Let's consider Sphere 1 with mass 3kg. The forces acting on Sphere 1 are:
1. Tension force (T) in the upward direction.
2. Weight force (m1g) in the downward direction.

The net force acting on Sphere 1 is given by:
Net force = T - m1g

According to Newton's second law of motion, the net force is equal to the product of mass and acceleration:
Net force = m1 * a1

Equating the net force and the product of mass and acceleration, we get:
T - m1g = m1 * a1 --------(1)

Acceleration of Sphere 2:
Now let's consider Sphere 2 with mass 4kg. The forces acting on Sphere 2 are:
1. Tension force (T) in the downward direction.
2. Weight force (m2g) in the downward direction.

The net force acting on Sphere 2 is given by:
Net force = m2g - T

Applying Newton's second law of motion, we have:
m2g - T = m2 * a2 --------(2)

Relative Acceleration of the System:
Since the two spheres are connected by a string, they have the same magnitude of acceleration but in opposite directions. Therefore, a1 = -a2.

Substituting a1 = -a2 in equation (1) and (2), we get:
T - m1g = m1 * (-a2) --------(3)
m2g - T = m2 * a2 --------(4)

Adding equations (3) and (4), we eliminate the tension force:
(m2g - T) + (T - m1g) = m1 * (-a2) + m2 * a2
m2g - m1g = (-m1 - m2) * a2
g(m2 - m1) = (m1 + m2) * a2

Finally, we can solve for the relative acceleration (a2):
a2 = (g(m2 - m1))/(m1 + m2)

Substituting the given values, we have:
a2 = (9.8(4 - 3))/(3 + 4)
a2 = 0.98 m/s²

Therefore, the relative acceleration of the system is 0.98 m/s².

20/7
correct??