Energy of springs connected in series and parallel

[songoku]

Homework Statement

Please see below

Relevant Equations

F = kx
$E = \frac{1}{2}kx^2$

For P, the force will be 2F and the spring constant will be $\frac{k}{2}$ so the extension will be 4 times, and the energy will be 8E but there are no options showing 8E

What is my mistake?

Thanks

 

[haruspex]

For P, the force will be 2F

How do you arrive at that?

 

[Orodruin]

The problem pretty clearly states that force is F, not 2F.

My hint for this problem would be to consider the forces on each constituent spring individually and then add up the energies. It makes things so much easier.

 

[songoku]

How do you arrive at that?

The problem pretty clearly states that force is F, not 2F.

My hint for this problem would be to consider the forces on each constituent spring individually and then add up the energies. It makes things so much easier.

I thought since there are two F in the diagrams, it means the force acting on the system is 2F. But it seems I misunderstood the question

 

[haruspex]

Tension is not a force, exactly, but a pair of equal and opposite forces. Likewise compression.
If a static rope is pulled by a force F at one end then it must also be pulled by F at the other.
What we say is the magnitude of the tension is a matter of convention, namely, that the tension is F. It could have been defined as 2F, but it isn’t.
In the more general case of a massive rope being pulled by unequal forces, F and F', the tension varies along the rope. At a small element of the rope, mass dm, the two forces differ by a.dm, where a is the acceleration of the rope.

 

[Orodruin]

It is even simpler than that.

Consider a mass $m$ hanging vertically from a fixed ceiling by means of an ideal spring. This is kind of the standard example of a spring being exposed to an external load.

Answer the following:
What is the tensile force in the spring?
What is the force with which the mass acts upon the spring with at thd bottom?
What is the force with which the ceiling acts upon the spring?
Can you draw a free body diagram of the spring?

 

[DrBanana]

It is even simpler than that.

Consider a mass $m$ hanging vertically from a fixed ceiling by means of an ideal spring. This is kind of the standard example of a spring being exposed to an external load.

Answer the following:
What is the tensile force in the spring?
What is the force with which the mass acts upon the spring with at thd bottom?
What is the force with which the ceiling acts upon the spring?
Can you draw a free body diagram of the spring?

I'm not OP but I'm having trouble with this as well. Here are my attempts at answers to your questions:

1. Assuming the spring constant is $k$ and its length is changed by $x$, then the tensile force is $kx$.

2. The spring is said to be ideal (massless) so it's nonsensical to ask what force acts on it?

3. This one I can't answer. If I had to guess I'd say it would be the weight of the object, plus $kx$ (because the string is pulling down from the ceiling).

4. Sorry for the low quality image

 

[Steve4Physics]

I'm not OP but I'm having trouble with this as well. Here are my attempts at answers to your questions:

1. Assuming the spring constant is k and its length is changed by x, then the tensile force is kx.

2. The spring is said to be ideal (massless) so it's nonsensical to ask what force acts on it?

3. This one I can't answer. If I had to guess I'd say it would be the weight of the object, plus kx (because the string is pulling down from the ceiling).

4. Sorry for the low quality image

Hi @DrBanana. There are some mistakes in your answers! Maybe you’d like to work through the following and then try answering @Orodruin ’s Post #6 questions again. Click on fuzzy ‘spoiler’ to reveal each answer below.

You have a piece of string. The string’s weight is negligible compared to other forces that will act on it; so we can treat the string as ‘massless’. You hold the string and pull it horizontal:
- your left hand exerts a force of 10N the left, on left end the string;
- your right hand exerts a force of 10N to the right, on the right end of the string.

Q1 What is the force exerted by the string on your left hand?

10N to the right (using Newton’s 3rd Law).

Q2 What is the force exerted by the string on your right hand?

10N to the left (using Newton’s 3rd Law).

Q3 List all force acting on the string and state the net (total) force.

10N to the left (the pull of your left hand).
10N to the right (the pull of your right hand).
Net force= 0 because (+10N) + (-10N) = 0.

Q4 What is the value of the tension (tensile force)) in the string?

Tension = 10N (because of how we define tension (tensile force) - see earlier posts.

Q5 Would the answer to Q4 be any different it the string had been vertical and we applied a 10N force up at the top and a 10N force down at the bottom?

No. The tension would still be 10N, The string’s weight is too small to make a difference. (A heavy string would make things more complicated so we don’t consider that at an introductory teaching level.).

Q If asked for a free-body diagram of the string, should the hands be included in the diagram?

No. Only the string and the forces acting on it (labelled arrows) should be on the free-body diagram.

Q7 Would using a spring (also of negligible weight) instead of the string make any difference to the tension?

No. The answer would still be the same. The spring’s tension would be 10N. But, in addition, to reach equilibrium the spring would have stretched by an amount x, such that kx = 10N where k is the spring constant.

EDIT: Speling korected.

 

[Orodruin]

1. Assuming the spring constant is $k$ and its length is changed by $x$, then the tensile force is $kx$.

You were not given the spring constant nor the extension of the spring.

2. The spring is said to be ideal (massless) so it's nonsensical to ask what force acts on it?

No. That an object is massless does not mean no forces can act on it. It means the net force must be zero.

3. This one I can't answer. If I had to guess I'd say it would be the weight of the object, plus $kx$ (because the string is pulling down from the ceiling).

Try Newton’s second law …

4. Sorry for the low quality image

View attachment 351470

This is not a free body diagram of the spring, as has been pointed out.

 

[DrBanana]

Thanks. Concerning the original problem, the correct answer is

Spoiler

D

, right?

 

[songoku]

It is even simpler than that.

Consider a mass $m$ hanging vertically from a fixed ceiling by means of an ideal spring. This is kind of the standard example of a spring being exposed to an external load.

Answer the following:
What is the tensile force in the spring?

mg

What is the force with which the mass acts upon the spring with at thd bottom?

mg

What is the force with which the ceiling acts upon the spring?

should be mg too because the spring is in equilibrium

Can you draw a free body diagram of the spring?

There are 2 forces: downwards from the mass and upwards from the ceiling, both have magnitude of mg

Is that correct?

Thanks. Concerning the original problem, the correct answer is

Spoiler

D

, right?

Yes

 

[Orodruin]

There are 2 forces: downwards from the mass and upwards from the ceiling, both have magnitude of mg

Is that correct?

Yes, that is correct. Now compare that free body diagram to the figures in your original post. (Just turn them 90 degrees)

 

[songoku]

I understand.

Thank you very much haruspex, Orodruin, DrBanana, Steve4Physics

 

[Alex Schaller]

I understand.

Thank you very much haruspex, Orodruin, DrBanana, Steve4Physics

P arrangement:
The force F acts along the string that links both springs. So, there would be a force F acting on Spring 1, and the same force acting on Spring 2.

Therefore, both springs would be elongated the same distance x, as just one spring would be elongated if a force F is applied to the spring.

So, each spring would have an energy E, and the P arrangement would have an energy equal 2 x E.

Conversely, and following a same line of thought, Q arrangement is going to end up with energy equal 0.5 x E.

You can deduct which is the correct answer to your homework.

Kind regards,
Alex