The blocks A and B shown in fig have masses M of A=5kg and M of B=4kg .The system is released from rest.The speed of B after A has travelled a distance 1m along the incline is

To determine the speed of block B after block A has traveled a distance of 1 meter along the incline, we can apply the principles of energy conservation and kinematics. Let's break down the problem step by step.

Understanding the System

We have two blocks: block A, which is on an incline, and block B, which is hanging vertically. The masses are given as:

• Mass of A (M_A) = 5 kg
• Mass of B (M_B) = 4 kg

When the system is released, block A will slide down the incline, causing block B to accelerate downward due to gravity. We need to find the speed of block B after block A has moved 1 meter down the incline.

Applying Energy Conservation

In this scenario, we can use the conservation of mechanical energy. The potential energy lost by block A as it moves down the incline will convert into kinetic energy for both blocks. The potential energy change for block A can be calculated as:

Potential Energy Change

The potential energy (PE) lost by block A when it descends 1 meter can be expressed as:

PE_A = M_A * g * h

Where:

• g = acceleration due to gravity (approximately 9.81 m/s²)
• h = vertical height change corresponding to the incline distance traveled

Assuming the incline angle is θ, the vertical height (h) can be calculated as:

h = d * sin(θ)

where d is the distance traveled along the incline (1 meter in this case).

Kinetic Energy of the System

The total kinetic energy (KE) gained by both blocks can be expressed as:

KE = 0.5 * M_A * v_A² + 0.5 * M_B * v_B²

Since block A and block B are connected, they will have the same speed (v) at the moment we are considering. Therefore, we can simplify the kinetic energy expression:

KE = 0.5 * (M_A + M_B) * v²

Setting Up the Equation

By applying the conservation of energy principle, we set the potential energy lost equal to the kinetic energy gained:

M_A * g * h = 0.5 * (M_A + M_B) * v²

Substituting h = d * sin(θ) and d = 1 m, we can solve for v:

Solving for Speed

Rearranging the equation gives us:

v² = (2 * M_A * g * h) / (M_A + M_B)

Now, substituting the values:

• M_A = 5 kg
• M_B = 4 kg
• g = 9.81 m/s²
• h = 1 m (assuming a vertical drop of 1 m for simplicity)

v² = (2 * 5 kg * 9.81 m/s² * 1 m) / (5 kg + 4 kg)

v² = (98.1 kg·m²/s²) / 9 kg

v² = 10.9 m²/s²

Taking the square root gives us:

v ≈ 3.30 m/s

Final Result

Thus, the speed of block B after block A has traveled a distance of 1 meter along the incline is approximately 3.30 m/s. This approach illustrates how energy conservation principles can be effectively applied to solve problems involving connected systems.