Hooke's Law and suspended objects

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Homework Help Overview

The problem involves Hooke's Law and the behavior of a spring when different masses are suspended from it. The original poster presents a scenario where a mass is suspended from a spring, and then questions how the spring behaves when two smaller masses are used instead.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to analyze the spring's stretch based on the weight of the masses and applies Hooke's Law. Some participants question the role of tension in the spring and how it relates to the stretch in both scenarios. Others explore the implications of having two masses and whether they both contribute to the spring's stretch.

Discussion Status

Participants are actively discussing the relationship between tension and stretch in the spring. There is an exploration of how the forces acting on the spring differ between the two configurations. Some guidance has been provided regarding the necessity of both masses in contributing to the spring's behavior, but no consensus has been reached on the interpretation of the problem.

Contextual Notes

There is a mention of a diagram that describes the setup, which may provide additional context. The discussion also reflects on the implications of removing one mass and how that affects the spring's tension and stretch.

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Homework Statement



A body of mass (m) is suspended from a spring with spring constant k in configuration (a) and the spring is stretched 0.1m. If two identical bodies of mass (m/2) are suspended from a spring with the same spring constant k in configuration (b), how much will the spring stretch? Explain your answer.

Homework Equations



F = -kx (Hooke's Law)

The Attempt at a Solution



Since in the first case the mass is at rest,
by mg - k(0.1) = 0, i calculated that k = 10mg

In the second case, since the bodies suspended have a mass of (m[tex]\div[/tex]2), their weight is halved too. Therefore the spring will stretch for 0.1[tex]\div[/tex]2 = 0.05m on each side.

i.e. It stretches for 0.1m (0.05m each on left and right)

However, according to the solution, the answer is 0.05m.
It is also stated that "if you have problems, substitute one of the two masses with a fixed hook; does the spring change its length?"

What difference does that make?

Attached is the diagram describing the setup
 

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Hint: The stretch of the spring is proportional to the tension. What's the tension in the spring in each scenario?
 
Doc Al said:
Hint: The stretch of the spring is proportional to the tension. What's the tension in the spring in each scenario?

Tension in the spring?

Case 1:
T = mg

Case 2:
T = mg/2
But it is pulled from both sides!
Do both blocks of objects with mass (m/2) cause the spring to stretch?

This is getting confusing...
 
annatar said:
Tension in the spring?

Case 1:
T = mg

Case 2:
T = mg/2
Good.
But it is pulled from both sides!
It better be! Hint: The spring is being pulled from both sides in both scenarios.

Do both blocks of objects with mass (m/2) cause the spring to stretch?
Sure. You need both blocks pulling to stretch the spring. Remove one, and the spring will go flying as the mass falls.
 
So...

In case 2:

One of the masses is responsible for preventing the spring from flying out,
while the other one causes the stretch?

Therefore stretch length = 0.1/2 m = 0.05 m (since weight causing the stretch is halved)
 
annatar said:
So...

In case 2:

One of the masses is responsible for preventing the spring from flying out,
while the other one causes the stretch?
Don't think of it that way. To stretch a (massless) spring you must pull it from both ends with the same force--that force equals the tension in the spring.

In case 1, the ceiling pulls up on one end while the mass pulls down on the other; both ends of the spring are pulled with a force equal to mg. The tension in the spring equals mg; it stretches accordingly.

In case 2, each end is pulled with a force equal to mg/2. The tension in the spring equals mg/2; it stretches accordingly.
 

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