Hooke's Law and suspended objects

In summary: Remember, the spring is massless. It can't have a mind of its own. The tension in the spring is the same throughout its length.In summary, a body of mass (m) suspended from a spring with spring constant k in configuration (a) causes the spring to stretch by 0.1m. In configuration (b), two identical bodies of mass (m/2) suspended from a spring with the same spring constant k will cause the spring to stretch by 0.05m on each side. The tension in the spring is proportional to the stretch and is equal to mg in configuration (a) and mg/2 in configuration (b). Both masses are responsible for causing the spring to stretch in
  • #1
annatar
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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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  • #2
Hint: The stretch of the spring is proportional to the tension. What's the tension in the spring in each scenario?
 
  • #3
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...
 
  • #4
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.
 
  • #5
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)
 
  • #6
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.
 

1. What is Hooke's Law?

Hooke's Law states that the force needed to extend or compress a spring is directly proportional to the distance the spring is stretched or compressed.

2. How does Hooke's Law apply to suspended objects?

Hooke's Law can also be applied to suspended objects, such as a hanging mass on a spring. The force of gravity pulling down on the mass is balanced by the force of the spring pushing upwards, creating a state of equilibrium.

3. What is the formula for Hooke's Law?

The formula for Hooke's Law is F = -kx, where F is the force applied to the spring, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

4. How does the spring constant affect the force applied to a spring?

The spring constant, represented by the variable k, determines the stiffness of the spring. A higher spring constant means a stiffer spring, which will require more force to stretch or compress it a certain distance.

5. Are there any limitations to Hooke's Law?

Yes, there are limitations to Hooke's Law. It only applies to objects that behave elastically, meaning they return to their original shape after being stretched or compressed. It also assumes that the force applied is not too large and that the spring is being stretched or compressed within its elastic limit.

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