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Two blocks of masses 8 kg and 12 kg are connected at the two ends of a light inextensible string. The string passes over a frictionless pulley. The acceleration of the system is
- a)g/4
- b)g/5
- c)g/8
- d)g/6
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Two blocks of masses 8 kg and 12 kg are connected at the two ends of a...
Let a be the common acceleration of the system and T be tension of the string. The equations of motion of two blocks are
T − 8g = 8a …(i)
and 12g −T = 12a
Adding (i) and (ii), we get
4g = 20a or a = g/5
T − 8g = 8a …(i)
and 12g −T = 12a
Adding (i) and (ii), we get
4g = 20a or a = g/5
Most Upvoted Answer
Two blocks of masses 8 kg and 12 kg are connected at the two ends of a...
Given information:
- Two blocks with masses 8 kg and 12 kg are connected by a light inextensible string.
- The string passes over a frictionless pulley.
To find:
- The acceleration of the system.
Let's solve this problem step by step:
1. Identify the forces acting on each block:
- Block 1 (8 kg): It experiences its weight (mg) acting downward and tension (T) acting upward.
- Block 2 (12 kg): It experiences its weight (mg) acting downward and tension (T) acting upward.
2. Apply Newton's second law to each block:
- Block 1: The net force acting on block 1 is the difference between the tension and the weight: T - mg.
- Block 2: The net force acting on block 2 is the difference between the weight and the tension: mg - T.
3. Set up the equations of motion:
- Block 1: T - mg = ma (equation 1)
- Block 2: mg - T = Ma (equation 2)
Here, a is the acceleration of block 1 and block 2, and M is the mass of block 2.
4. Substitute the given values:
- Block 1: T - (8 kg)(9.8 m/s^2) = (8 kg)a
- Block 2: (12 kg)(9.8 m/s^2) - T = (12 kg)a
5. Solve the equations simultaneously:
- Add the two equations: T - (8 kg)(9.8 m/s^2) + (12 kg)(9.8 m/s^2) - T = (8 kg + 12 kg)a
- Simplify: (12 kg)(9.8 m/s^2) - (8 kg)(9.8 m/s^2) = (20 kg)a
- Calculate: (12 kg)(9.8 m/s^2) - (8 kg)(9.8 m/s^2) = (20 kg)a
- Simplify: (4 kg)(9.8 m/s^2) = (20 kg)a
- Solve for a: a = (4 kg)(9.8 m/s^2) / (20 kg)
6. Calculate the acceleration:
- Plug in the values: a = 3.92 m/s^2
Therefore, the acceleration of the system is 3.92 m/s^2.
Hence, the correct answer is option B: g/5.
- Two blocks with masses 8 kg and 12 kg are connected by a light inextensible string.
- The string passes over a frictionless pulley.
To find:
- The acceleration of the system.
Let's solve this problem step by step:
1. Identify the forces acting on each block:
- Block 1 (8 kg): It experiences its weight (mg) acting downward and tension (T) acting upward.
- Block 2 (12 kg): It experiences its weight (mg) acting downward and tension (T) acting upward.
2. Apply Newton's second law to each block:
- Block 1: The net force acting on block 1 is the difference between the tension and the weight: T - mg.
- Block 2: The net force acting on block 2 is the difference between the weight and the tension: mg - T.
3. Set up the equations of motion:
- Block 1: T - mg = ma (equation 1)
- Block 2: mg - T = Ma (equation 2)
Here, a is the acceleration of block 1 and block 2, and M is the mass of block 2.
4. Substitute the given values:
- Block 1: T - (8 kg)(9.8 m/s^2) = (8 kg)a
- Block 2: (12 kg)(9.8 m/s^2) - T = (12 kg)a
5. Solve the equations simultaneously:
- Add the two equations: T - (8 kg)(9.8 m/s^2) + (12 kg)(9.8 m/s^2) - T = (8 kg + 12 kg)a
- Simplify: (12 kg)(9.8 m/s^2) - (8 kg)(9.8 m/s^2) = (20 kg)a
- Calculate: (12 kg)(9.8 m/s^2) - (8 kg)(9.8 m/s^2) = (20 kg)a
- Simplify: (4 kg)(9.8 m/s^2) = (20 kg)a
- Solve for a: a = (4 kg)(9.8 m/s^2) / (20 kg)
6. Calculate the acceleration:
- Plug in the values: a = 3.92 m/s^2
Therefore, the acceleration of the system is 3.92 m/s^2.
Hence, the correct answer is option B: g/5.
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Two blocks of masses 8 kg and 12 kg are connected at the two ends of a light inextensible string. The string passes over a frictionless pulley. The acceleration of the system isa)g/4b)g/5c)g/8d)g/6Correct answer is option 'B'. Can you explain this answer?
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