The masses of the bodies A and B in the given fig ( atached image) are 20 kg and 10 kg respectively ,initially at rest on the floor and are connected by aweightless string passing oveta weight less and frictionless pulley an upwatd force F is applied to the pulley 1) the acc of a1 of body A and a2 of B when F is 98N,96N, 394N respectively

To determine the accelerations of bodies A and B when different forces are applied to the pulley, we can use Newton's second law of motion. This law states that the force acting on an object is equal to the mass of that object multiplied by its acceleration (F = ma). In this scenario, we have two bodies connected by a string over a pulley, and we need to analyze the forces acting on each body when an upward force is applied to the pulley.

Understanding the System

We have two bodies: A with a mass of 20 kg and B with a mass of 10 kg. The system is influenced by the gravitational force acting on both bodies and the applied force F on the pulley. The gravitational force can be calculated using the formula:

• Weight of A (W_A) = m_A * g = 20 kg * 9.81 m/s² = 196.2 N
• Weight of B (W_B) = m_B * g = 10 kg * 9.81 m/s² = 98.1 N

Net Force Calculation

The net force acting on the system when the upward force F is applied can be calculated by considering the forces acting on both bodies. The net force (F_net) can be expressed as:

F_net = F - (W_A - W_B)

Now, let's calculate the accelerations for the different values of F: 98 N, 96 N, and 394 N.

Case 1: F = 98 N

For F = 98 N, we calculate the net force:

F_net = 98 N - (196.2 N - 98.1 N) = 98 N - 98.1 N = -0.1 N

Since the net force is negative, the system will not accelerate upwards. Instead, it will accelerate downwards. The total mass of the system is:

m_total = m_A + m_B = 20 kg + 10 kg = 30 kg

Using F = ma, we find the acceleration:

a = F_net / m_total = -0.1 N / 30 kg = -0.00333 m/s²

Case 2: F = 96 N

Now, for F = 96 N:

F_net = 96 N - (196.2 N - 98.1 N) = 96 N - 98.1 N = -2.1 N

Again, the system accelerates downwards:

a = F_net / m_total = -2.1 N / 30 kg = -0.07 m/s²

Case 3: F = 394 N

Finally, for F = 394 N:

F_net = 394 N - (196.2 N - 98.1 N) = 394 N - 98.1 N = 295.9 N

This time, the net force is positive, indicating that the system accelerates upwards:

a = F_net / m_total = 295.9 N / 30 kg = 9.863 m/s²

Summary of Accelerations

To summarize the accelerations for each case:

• When F = 98 N, acceleration a ≈ -0.00333 m/s² (downwards)
• When F = 96 N, acceleration a ≈ -0.07 m/s² (downwards)
• When F = 394 N, acceleration a ≈ 9.863 m/s² (upwards)

This analysis shows how the applied force affects the motion of the system, demonstrating the principles of dynamics in action. If you have any further questions or need clarification on any part, feel free to ask!