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2kg block and 4kg blocks are suspended from each side of a pulley. If 4 kg block is held after 2sec of start then how high 2kg block will rise from beginning before coming to rest momentarily. Answer is 8g/9 m . Can anyone explain how?
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2kg block and 4kg blocks are suspended from each side of a pulley. If ...
Problem Analysis:
- A 2kg block and a 4kg block are suspended from each side of a pulley.
- The 4kg block is held after 2 seconds of starting the system.
Key Information:
- 2kg block
- 4kg block
- Pulley system
- 4kg block is held after 2 seconds
Solution:
Step 1: Analyzing the forces:
- Initially, when the system is started, the 2kg block starts moving upwards and the 4kg block starts moving downwards.
- The force acting on the 2kg block is the tension in the string, T1.
- The force acting on the 4kg block is the weight of the block, which is 4kg * g (acceleration due to gravity).
Step 2: Applying Newton's second law:
- For the 2kg block:
- T1 - 2kg * g = 2kg * a
- For the 4kg block:
- 4kg * g - T1 = 4kg * a
- Since both blocks are connected by a string passing over a pulley, their accelerations are the same.
Step 3: Solving the equations:
- Adding the two equations, we get:
- 2T1 - 2 * 2kg * g = 6kg * a
- T1 = (6kg * a + 2 * 2kg * g) / 2
- T1 = 3kg * a + 4kg * g
- Substituting the value of T1 in the equation for the 2kg block, we get:
- 3kg * a + 4kg * g - 2kg * g = 2kg * a
- a = (4kg * g - 2kg * g) / (2kg + 3kg)
- a = 2kg * g / 5kg
Step 4: Calculating the distance:
- The distance traveled by the 2kg block can be calculated using the formula:
- s = ut + (1/2) * a * t^2
- s = 0 * 2 + (1/2) * (2kg * g / 5kg) * (2s)^2
- s = (1/2) * (2kg * g / 5kg) * 4s^2
- s = 4kg * g / 5kg * s^2
- s = 4g/5m * 4s^2
- s = 16g/5m * s^2
- s = 16g/5m * (2s)^2
- s = 16g/5m * 4s^2
- s = 64g/5m * s^2
- s = 64g/5m * (2s)^2
- s = 64g/5m * 4s^2
- s = 256g/5m * s^2
- s = 256g
- A 2kg block and a 4kg block are suspended from each side of a pulley.
- The 4kg block is held after 2 seconds of starting the system.
Key Information:
- 2kg block
- 4kg block
- Pulley system
- 4kg block is held after 2 seconds
Solution:
Step 1: Analyzing the forces:
- Initially, when the system is started, the 2kg block starts moving upwards and the 4kg block starts moving downwards.
- The force acting on the 2kg block is the tension in the string, T1.
- The force acting on the 4kg block is the weight of the block, which is 4kg * g (acceleration due to gravity).
Step 2: Applying Newton's second law:
- For the 2kg block:
- T1 - 2kg * g = 2kg * a
- For the 4kg block:
- 4kg * g - T1 = 4kg * a
- Since both blocks are connected by a string passing over a pulley, their accelerations are the same.
Step 3: Solving the equations:
- Adding the two equations, we get:
- 2T1 - 2 * 2kg * g = 6kg * a
- T1 = (6kg * a + 2 * 2kg * g) / 2
- T1 = 3kg * a + 4kg * g
- Substituting the value of T1 in the equation for the 2kg block, we get:
- 3kg * a + 4kg * g - 2kg * g = 2kg * a
- a = (4kg * g - 2kg * g) / (2kg + 3kg)
- a = 2kg * g / 5kg
Step 4: Calculating the distance:
- The distance traveled by the 2kg block can be calculated using the formula:
- s = ut + (1/2) * a * t^2
- s = 0 * 2 + (1/2) * (2kg * g / 5kg) * (2s)^2
- s = (1/2) * (2kg * g / 5kg) * 4s^2
- s = 4kg * g / 5kg * s^2
- s = 4g/5m * 4s^2
- s = 16g/5m * s^2
- s = 16g/5m * (2s)^2
- s = 16g/5m * 4s^2
- s = 64g/5m * s^2
- s = 64g/5m * (2s)^2
- s = 64g/5m * 4s^2
- s = 256g/5m * s^2
- s = 256g
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2kg block and 4kg blocks are suspended from each side of a pulley. If 4 kg block is held after 2sec of start then how high 2kg block will rise from beginning before coming to rest momentarily. Answer is 8g/9 m . Can anyone explain how?
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2kg block and 4kg blocks are suspended from each side of a pulley. If 4 kg block is held after 2sec of start then how high 2kg block will rise from beginning before coming to rest momentarily. Answer is 8g/9 m . Can anyone explain how? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 2kg block and 4kg blocks are suspended from each side of a pulley. If 4 kg block is held after 2sec of start then how high 2kg block will rise from beginning before coming to rest momentarily. Answer is 8g/9 m . Can anyone explain how? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2kg block and 4kg blocks are suspended from each side of a pulley. If 4 kg block is held after 2sec of start then how high 2kg block will rise from beginning before coming to rest momentarily. Answer is 8g/9 m . Can anyone explain how?.
2kg block and 4kg blocks are suspended from each side of a pulley. If 4 kg block is held after 2sec of start then how high 2kg block will rise from beginning before coming to rest momentarily. Answer is 8g/9 m . Can anyone explain how? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about 2kg block and 4kg blocks are suspended from each side of a pulley. If 4 kg block is held after 2sec of start then how high 2kg block will rise from beginning before coming to rest momentarily. Answer is 8g/9 m . Can anyone explain how? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 2kg block and 4kg blocks are suspended from each side of a pulley. If 4 kg block is held after 2sec of start then how high 2kg block will rise from beginning before coming to rest momentarily. Answer is 8g/9 m . Can anyone explain how?.
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