Pulleys
In an Atwood machine m1 = 6 kg, m2 = 5 kg. Acceleration is (g = 10 m/s²):
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Solution
(1/11) g
For an Atwood machine consisting of two masses m1 and m2 connected by a light string over a frictionless pulley (m1>m2), the acceleration a is: a=m1+m2(m1−m2)g
- Numerator: The net driving force is the difference in weights: m1g−m2g=(6−5)g=1g.
- Denominator: The total mass being accelerated is m1+m2=6+5=11 kg. Substituting: a=111g=111g. In terms of value, a≈0.91 m/s2.
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About This Question
- Subject
- physics
- Chapter
- laws of motion
- Topic
- pulleys
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
(1/11) g
For an Atwood machine consisting of two masses m1 and m2 connected by a light string over a frictionless pulley (m1>m2), the acceleration a is: a=m1+m2(m1−m2)g
- Numerator: The net driving force is the difference in weights: m1g−m2g=(6−5)g=1g.
- Denominator: The total mass being accelerated is m1+m2=6+5=11 kg. Substituting: a=111g=111g. In terms of value, a≈0.91 m/s2.
This medium difficulty physics question is from the chapter laws of motion, covering the topic of pulleys. It appeared in the 2025 exam.
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