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A ball of mass m is attached to the lower end of light vertical spring of force constant k. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstreched) length, comes to rest again after descending through a distance x.
- a)x = mg/k
- b)x = 2 mg/k
- c)The ball will have no acceleration at the position where it has descended through x/2.
- d)The ball will have an upward acceleration equal to g at its lowermost position.
Correct answer is option 'B,C,D'. Can you explain this answer?
Verified Answer
A ball of mass m is attached to the lower end of light vertical spring...
The ball is at rest, it has maximum potential energy. When the ball is released from rest with the spring at its normal (unstretched) length it loses some potential energy and energy of spring increases. Hence, loss in potential energy of the ball is equal to gain in potential energy of spring.
∴mgx=1/2kx2
∴x=2mg/k
Also, for x′=x/2,
kx′=mg i.e. forces are equal thus, the ball will have no acceleration at the position where it has descended through 2x.
And when ball is at lowermost position, the spring force will be
kx=2mg
Hence, the ball will have an upward acceleration equal to g at its lowermost position.
∴mgx=1/2kx2
∴x=2mg/k
Also, for x′=x/2,
kx′=mg i.e. forces are equal thus, the ball will have no acceleration at the position where it has descended through 2x.
And when ball is at lowermost position, the spring force will be
kx=2mg
Hence, the ball will have an upward acceleration equal to g at its lowermost position.
Most Upvoted Answer
A ball of mass m is attached to the lower end of light vertical spring...
Explanation:
1. Initial conditions:
- The ball is attached to a light vertical spring of force constant k.
- The spring is fixed at its upper end.
- The ball is released from rest with the spring at its normal (unstretched) length.
2. Forces acting on the ball:
- Gravity force (mg) acts downward on the ball.
- Spring force (F_s) acts upward on the ball due to the compression of the spring.
3. Forces at different positions:
a) At the highest point of motion (rest):
- At this point, the ball is at its maximum displacement from the equilibrium position.
- The net force on the ball is zero because the spring force and the gravitational force cancel each other out.
- Therefore, the ball will have no acceleration at this position.
b) At the midpoint of motion (descended through x/2):
- At this point, the ball has descended through a distance x/2.
- The net force on the ball is upwards and equal to the spring force.
- The spring force is given by Hooke's law: F_s = -kx, where x is the displacement of the ball.
- The acceleration of the ball is given by Newton's second law: F_net = ma.
- Therefore, the ball will have an upward acceleration at this position.
c) At the lowermost point of motion (descended through x):
- At this point, the ball has descended through a distance x.
- The net force on the ball is upwards and equal to the spring force.
- The gravitational force is also acting downwards.
- The acceleration of the ball is given by Newton's second law: F_net = ma.
- The net force is given by F_net = F_s - mg, where mg is the gravitational force.
- Since the ball is at rest at this position, the net force must be zero.
- Therefore, the spring force must be equal to the gravitational force, i.e., F_s = mg.
- This implies that x = 2mg/k.
Conclusion:
From the above analysis, we can conclude that:
- The ball will have no acceleration at the position where it has descended through x/2.
- The ball will have an upward acceleration equal to g at its lowermost position.
- The ball will have descended through a distance x = 2mg/k.
Therefore, the correct answer is options B, C, and D.
1. Initial conditions:
- The ball is attached to a light vertical spring of force constant k.
- The spring is fixed at its upper end.
- The ball is released from rest with the spring at its normal (unstretched) length.
2. Forces acting on the ball:
- Gravity force (mg) acts downward on the ball.
- Spring force (F_s) acts upward on the ball due to the compression of the spring.
3. Forces at different positions:
a) At the highest point of motion (rest):
- At this point, the ball is at its maximum displacement from the equilibrium position.
- The net force on the ball is zero because the spring force and the gravitational force cancel each other out.
- Therefore, the ball will have no acceleration at this position.
b) At the midpoint of motion (descended through x/2):
- At this point, the ball has descended through a distance x/2.
- The net force on the ball is upwards and equal to the spring force.
- The spring force is given by Hooke's law: F_s = -kx, where x is the displacement of the ball.
- The acceleration of the ball is given by Newton's second law: F_net = ma.
- Therefore, the ball will have an upward acceleration at this position.
c) At the lowermost point of motion (descended through x):
- At this point, the ball has descended through a distance x.
- The net force on the ball is upwards and equal to the spring force.
- The gravitational force is also acting downwards.
- The acceleration of the ball is given by Newton's second law: F_net = ma.
- The net force is given by F_net = F_s - mg, where mg is the gravitational force.
- Since the ball is at rest at this position, the net force must be zero.
- Therefore, the spring force must be equal to the gravitational force, i.e., F_s = mg.
- This implies that x = 2mg/k.
Conclusion:
From the above analysis, we can conclude that:
- The ball will have no acceleration at the position where it has descended through x/2.
- The ball will have an upward acceleration equal to g at its lowermost position.
- The ball will have descended through a distance x = 2mg/k.
Therefore, the correct answer is options B, C, and D.
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A ball of mass m is attached to the lower end of light vertical spring of force constant k. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstreched) length, comes to rest again after descending through a distance x.a)x = mg/kb)x = 2 mg/kc)The ball will have no acceleration at the position where it has descended through x/2.d)The ball will have an upward acceleration equal to g at its lowermost position.Correct answer is option 'B,C,D'. Can you explain this answer?
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A ball of mass m is attached to the lower end of light vertical spring of force constant k. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstreched) length, comes to rest again after descending through a distance x.a)x = mg/kb)x = 2 mg/kc)The ball will have no acceleration at the position where it has descended through x/2.d)The ball will have an upward acceleration equal to g at its lowermost position.Correct answer is option 'B,C,D'. Can you explain this answer? for Airforce X Y / Indian Navy SSR 2025 is part of Airforce X Y / Indian Navy SSR preparation. The Question and answers have been prepared according to the Airforce X Y / Indian Navy SSR exam syllabus. Information about A ball of mass m is attached to the lower end of light vertical spring of force constant k. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstreched) length, comes to rest again after descending through a distance x.a)x = mg/kb)x = 2 mg/kc)The ball will have no acceleration at the position where it has descended through x/2.d)The ball will have an upward acceleration equal to g at its lowermost position.Correct answer is option 'B,C,D'. Can you explain this answer? covers all topics & solutions for Airforce X Y / Indian Navy SSR 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball of mass m is attached to the lower end of light vertical spring of force constant k. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstreched) length, comes to rest again after descending through a distance x.a)x = mg/kb)x = 2 mg/kc)The ball will have no acceleration at the position where it has descended through x/2.d)The ball will have an upward acceleration equal to g at its lowermost position.Correct answer is option 'B,C,D'. Can you explain this answer?.
A ball of mass m is attached to the lower end of light vertical spring of force constant k. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstreched) length, comes to rest again after descending through a distance x.a)x = mg/kb)x = 2 mg/kc)The ball will have no acceleration at the position where it has descended through x/2.d)The ball will have an upward acceleration equal to g at its lowermost position.Correct answer is option 'B,C,D'. Can you explain this answer? for Airforce X Y / Indian Navy SSR 2025 is part of Airforce X Y / Indian Navy SSR preparation. The Question and answers have been prepared according to the Airforce X Y / Indian Navy SSR exam syllabus. Information about A ball of mass m is attached to the lower end of light vertical spring of force constant k. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstreched) length, comes to rest again after descending through a distance x.a)x = mg/kb)x = 2 mg/kc)The ball will have no acceleration at the position where it has descended through x/2.d)The ball will have an upward acceleration equal to g at its lowermost position.Correct answer is option 'B,C,D'. Can you explain this answer? covers all topics & solutions for Airforce X Y / Indian Navy SSR 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A ball of mass m is attached to the lower end of light vertical spring of force constant k. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal (unstreched) length, comes to rest again after descending through a distance x.a)x = mg/kb)x = 2 mg/kc)The ball will have no acceleration at the position where it has descended through x/2.d)The ball will have an upward acceleration equal to g at its lowermost position.Correct answer is option 'B,C,D'. Can you explain this answer?.
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