Consider a pulley having 0.6kg and 0.3 on either side.the larger mass is stopped for a moment of 2 seconds after the system is set into motion.find the time elapsed before the string is tight again?

To solve this problem, we need to analyze the motion of the pulley system and how the forces interact when one mass is momentarily stopped. We have two masses: 0.6 kg on one side and 0.3 kg on the other. When the system is set into motion, the heavier mass (0.6 kg) will accelerate downward while the lighter mass (0.3 kg) will accelerate upward. Let's break this down step by step.

Understanding the Forces at Play

Initially, when the system is released, the gravitational force acting on each mass will dictate their motion. The force due to gravity can be calculated using the formula:

Weight (W) = mass (m) × gravitational acceleration (g)

Here, g is approximately 9.81 m/s².

Calculating the Forces

• For the 0.6 kg mass: W₁ = 0.6 kg × 9.81 m/s² = 5.886 N
• For the 0.3 kg mass: W₂ = 0.3 kg × 9.81 m/s² = 2.943 N

Net Force and Acceleration

The net force acting on the system when both masses are in motion can be calculated as follows:

Net Force (F_net) = W₁ - W₂

Substituting the values:

F_net = 5.886 N - 2.943 N = 2.943 N

Next, we can find the acceleration (a) of the system using Newton's second law:

F_net = (m₁ + m₂) × a

Here, m₁ is 0.6 kg and m₂ is 0.3 kg:

2.943 N = (0.6 kg + 0.3 kg) × a

2.943 N = 0.9 kg × a

Thus, a = 2.943 N / 0.9 kg ≈ 3.26 m/s².

Time Calculation After Stopping the Heavier Mass

When the 0.6 kg mass is stopped for 2 seconds, the 0.3 kg mass will continue to rise during this time. We need to calculate how far the 0.3 kg mass moves upward in those 2 seconds:

Using the formula for distance (d) under constant acceleration:

d = 0.5 × a × t²

Substituting the values:

d = 0.5 × 3.26 m/s² × (2 s)² = 0.5 × 3.26 m/s² × 4 s² = 6.52 m.

Re-establishing the Tension in the String

Once the 0.6 kg mass is released again, it will fall the same distance that the 0.3 kg mass has risen to re-establish tension in the string. The time it takes for the 0.6 kg mass to fall this distance can be calculated using the same distance formula:

We need to find the time (t) it takes to fall 6.52 m under the acceleration of 9.81 m/s²:

Using the formula:

d = 0.5 × g × t²

Rearranging gives:

t² = 2d/g

Substituting the values:

t² = 2 × 6.52 m / 9.81 m/s² ≈ 1.33 s²

Thus, t ≈ √1.33 s² ≈ 1.15 s.

Final Time Calculation

Now, to find the total time elapsed before the string is tight again, we add the 2 seconds that the heavier mass was stopped to the time it took for it to fall:

Total Time = Time Stopped + Time to Fall

Total Time = 2 s + 1.15 s ≈ 3.15 s.

In summary, the time elapsed before the string is tight again is approximately 3.15 seconds. This approach illustrates how the dynamics of the pulley system work and how we can calculate the effects of stopping one mass on the overall motion of the system.