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An ideal spring with spring-constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is
- a)4 Mg/k
- b)2 Mg/k
- c)Mg/k
- d)Mg/2k
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
An ideal spring with spring-constant k is hung from the ceiling and a ...
Once the mass is released, the spring would extend Mg/k length where it comes to an equilibrium, but by then the mass would have gained some speed due to which the spring will be stretched Mg/k length more and then start a damped oscillation. Thus maximum extension by spring is 2Mg/k length.
Most Upvoted Answer
An ideal spring with spring-constant k is hung from the ceiling and a ...
**Explanation:**
The extension of the spring can be determined using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
**Hooke's Law:**
F = -kx
where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position.
When the block is released, it will start accelerating downwards due to the force of gravity. As the block accelerates downwards, the spring will stretch and exert an upward force on the block.
At the maximum extension of the spring, the gravitational force pulling the block downwards will be balanced by the upward force exerted by the spring.
**At maximum extension, the following equilibrium condition can be written:**
mg = kx
where m is the mass of the block, g is the acceleration due to gravity, and x is the maximum extension of the spring.
**Solving for x, we get:**
x = mg/k
Comparing this equation with the given options, it can be seen that the correct answer is option **(b) 2Mg/k**.
The extension of the spring can be determined using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.
**Hooke's Law:**
F = -kx
where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position.
When the block is released, it will start accelerating downwards due to the force of gravity. As the block accelerates downwards, the spring will stretch and exert an upward force on the block.
At the maximum extension of the spring, the gravitational force pulling the block downwards will be balanced by the upward force exerted by the spring.
**At maximum extension, the following equilibrium condition can be written:**
mg = kx
where m is the mass of the block, g is the acceleration due to gravity, and x is the maximum extension of the spring.
**Solving for x, we get:**
x = mg/k
Comparing this equation with the given options, it can be seen that the correct answer is option **(b) 2Mg/k**.
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Community Answer
An ideal spring with spring-constant k is hung from the ceiling and a ...
Since F=1/2kx for spring.
mg=1/2 kx
X=2mg/k
mg=1/2 kx
X=2mg/k
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An ideal spring with spring-constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring isa)4 Mg/kb)2 Mg/kc)Mg/kd)Mg/2kCorrect answer is option 'B'. Can you explain this answer?
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