Conservation of mechanical energy (no friction) - doesn't work in this case?

In summary: The equation for the "relaxed" position is:E = -kxThe equation for the initial compressed position is:E = -hx + kyThe equation for the ball losing contact with the spring is:E = -ky
  • #1
Femme_physics
Gold Member
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Conservation of mechanical energy (no friction) -- doesn't work in this case?

Homework Statement



http://img143.imageshack.us/img143/3181/springythingy.jpg

In the drawing is described a lab device designed to eject balls upwards. The balls are ejected as a result from the spiral spring being compressed initially. In relaxed position to uppermost part of the spring is in line with the uppermost part of the device.

Given:

Ball weight = 0.5 [N]
Spring's hardness coeffecient = 3 N/cm
Initial compression of the spring = 12 cm


Calculate

A) The speed in which the ball is ejected from the device
B) The height H which the ball rises to

Ignore friction and air drag.


Comment: The cumulative energy in the elastic force of a spiral spring on runway h equals

http://img818.imageshack.us/img818/5626/echyj.jpg


The Attempt at a Solution



I appear to have 2 unknowns in conservation of mechanical energy. I was told that when there's no friction I can use it to get the solution. What am I missing?

http://img818.imageshack.us/img818/8741/unnnnnnnnns.jpg
 
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  • #2


To make things clear, could you please list all variables (height of the ball and its velocity) and the energy terms
1. when the spring is compressed (initial state)
2 when the spring is relaxed
3 when the ball is at maximum height.

ehild
 
  • #3


Hi Fp! Glad to see you're still around! :smile:

Edited: as ehild said, you should break the problem down.
 
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  • #4


Hi ehild, ILS! :smile:

Still around? When did I go anywhere?!?

To make things clear, could you please list all variables (height of the ball and its velocity) and the energy terms

What do you mean by "height of the ball"? I'm not given the radius
1. when the spring is compressed (initial state)
2 when the spring is relaxed

The compression is h1 -- The initial position is when the spring is at the height of the device. This says so in the question
3 when the ball is at maximum height.

That would be "H" in my formula, but again I have 2 unknowns in the conservation of mechanical energy
 
  • #5


Femme_physics said:
What do you mean by "height of the ball"? I'm not given the radius
Sorry, I meant the position of the ball. At what height is it.

Femme_physics said:
The compression is h1 -- The initial position is when the spring is at the height of the device. This says so in the question
Sorry, the question says that the spring has been compressed initially. Before that, it was not compressed, but then somebody came and pressed it, but that somebody and his/her force is out of question.
So initially the spring is compressed. Where is the ball, what its its velocity and what are the energy terms (kinetic, gravitational and elastic) ?

ehild
 
  • #6


Femme_physics said:
Hi ehild, ILS! :smile:

Still around? When did I go anywhere?!?

Huh? :rolleyes:
I didn't say you went anywhere. I'm glad you're around! :smile:


Femme_physics said:
The compression is h1 -- The initial position is when the spring is at the height of the device. This says so in the question

Huh? I don't read that anywhere in the question?
 
  • #7
Hi Femme_physics! :smile:

Each part should only involve one equation. :confused:

Can you write out the equations you're using, with just symbols instead of numbers, so it's easy to check?
 
  • #8


Sorry, I meant the position of the ball. At what height is it.

I figured I can just define the height of the ball as my zero point. Can't I?

Sorry, the question says that the spring has been compressed initially. Before that, it was not compressed, but then somebody came and pressed it, but that somebody and his/her force is out of question.
So initially the spring is compressed. Where is the ball, what its its velocity and what are the energy terms (kinetic, gravitational and elastic) ?

ehild

Well, it's potential energy in the form of elastic energy

I didn't say you went anywhere. I'm glad you're around!

Ah, not "still" around, but just "around" :smile: Always planning to be! :D

Huh? I don't read that anywhere in the question?

Sorry, I didn't mean the "initial" position but the "relaxed" position of the spring
 
  • #9


tiny-tim said:
Hi Femme_physics! :smile:

Each part should only involve one equation. :confused:

Can you write out the equations you're using, with just symbols instead of numbers, so it's easy to check?

I was using this:


http://img845.imageshack.us/img845/616/mgh.jpg [/QUOTE]


I just didn't have enough room to do it all in one line so I used two lines
 
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  • #10


Femme_physics said:
Sorry, I didn't mean the "initial" position but the "relaxed" position of the spring

As I read it, the "relaxed" position is not the initial position.
Initially the spring is compressed.
Then it is released and uncompresses into its relaxed position.
After that the ball loses contact with the spring and jumps up.

Could you write the equation you just mentioned, applied to the initial compressed position and the relaxed position, just before the ball loses contact with the spring?
 
  • #11
ok, then i can only see one unknown at a time …

if you know h2 = 0, the only unknown is v2

if you know v2 = 0, the only unknown is h2 :wink:​
 
  • #12


As I read it, the "relaxed" position is not the initial position.
Initially the spring is compressed.
Then it is released and uncompresses into its relaxed position.
After that the ball loses contact with the spring and jumps up.

Yes that's why I said sorry :)
Could you write the equation you just mentioned, applied to the initial compressed position and the relaxed position, just before the ball loses contact with the spring?
I thought I did at my attempt at the solution. Should I make Vinitial to be 0? I was wondering that, because indeed H is my only unknown.

ok, then i can only see one unknown at a time …
if you know h2 = 0, the only unknown is v2

if you know v2 = 0, the only unknown is h2

Is V1 also 0? same question as above :smile:
 
  • #13


Femme_physics said:
Is V1 also 0? same question as above :smile:

Aha!
I think I just figured out what you did! :smile:

Yes, if you set V1 to 0, which it is, you'll only have 1 unknown left, which is H.
 
  • #14


Your life would be much easier with some systematic work. Let y the height where the ball is.

I. At the initial position of the ball: y=0, v=0, string compressed. Energy: (KE+PE+elastic)

II. relaxed state of the spring:
Ball at y = h1, v=v1, spring relaxed. Energy (KE + PE+elastic)

The ball lost contact with the spring.

III. At the highest poition y=H, v=0 no spring. Energy KE +PE.

The energy is the same at all positions, so you have one independent equation for v1 (EI=EII) and one for H (E1=EIII). ehild
 
  • #15
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  • #16


Femme_physics said:
Ah, I'm an idiot! I pretty much wrote this solution before I posted the problem but I screwed up the math, but I thought this might be the case!

Thanks for setting me straight! :)

Hmm, so you have a height H = 0.0432 m measured from the compressed position of the spring.
And the spring is compressed by h1 = 0.12 m.
What do you visual cues tell you about this?
Did you check all your units?
Femme_physics said:
Now Vo should be a piece a cake. I think.

But I got the wrong Vo according to the answers. It should be:
9.08 m/sec

What did you use for delta y exactly?
 
  • #17


Check the unit of the stiffness constant.

ehild
 
  • #18


I like Serena said:
Hmm, so you have a height H = 0.0432 m measured from the compressed position of the spring.
And the spring is compressed by h1 = 0.12 m.
What do you visual cues tell you about this?
Did you check all your units?




What did you use for delta y exactly?

I used 0.0432 meters for delta y

I always use meters, and since 12 cm is 0.12 meters, that's what I plugged in.



Check the unit of the stiffness constant.


That'll be 3 N/m


(heh, you said "stiffness" *chuckles*)
 
  • #19


Sorry I meant the spring constant that you called "hardness coefficient". Read its value in the text of the problem, please. And you do it very well if you convert everything to SI units. Do with the spring constant, too.

ehild
 
  • #20


Femme_physics said:
I always use meters, and since 12 cm is 0.12 meters, that's what I plugged in.

What I meant was that the spring is compressed 12 cm, and when released it apparently comes up 4.32 cm.
That's not even out of the device!
At that point the spring is still compressed.
That can' t be right! :confused:


Femme_physics said:
That'll be 3 N/m

Errr, no?
 
  • #21


Oh, I see. it's 3 N/cm :smile:

So now
3cm is 0.03m so I'll now plug everything in meters

And now since I converted everything to meters, result should be in meters

http://img98.imageshack.us/img98/5536/hequals.jpg

Which means I'm more off than I were before! :(
 
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  • #22


[tex]3 \ N/cm = 3 \ \frac N {cm} = 3 \cdot \frac N {0.01 m} = ...[/tex]
:smile:
 
  • #23


Do you say that 3 N/cm = 0.03 N/m? That you need less force when stretching by one m than stretching it by 1cm?

ehild
 
  • #24


Do you say that 3 N/cm = 0.03 N/m? That you need less force when stretching by one m than stretching it by 1cm?

Ah, I see the logic.

I like Serena said:
[tex]3 \ N/cm = 3 \ \frac N {cm} = 3 \cdot \frac N {0.01 m} = ...[/tex]
:smile:

Now if I that it turns out 300 N/m

Then now I do

http://img851.imageshack.us/img851/6400/hnew.jpg

Still not the right decimal point. Did I miss anything more?
 
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  • #25


Femme_physics said:
Still not the right decimal point. Did I miss anything more?

Yes. Calculate again.


ehild
 
  • #27


Femme_physics said:
:shy:

:blushing:

Ah! Great :smile:

:rolleyes:


Femme_physics said:
But my Vi is a bit off... anything wrong here?

Yes, your delta y is off.
From where to where should delta y be exactly?
 
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  • #28


Try to check what you have written before posting. The solution of your equation is Vi=0.ehild
 
  • #29


No no, I'm looking for ejected velocity, so I see what ILS is saying.

It's the point where the ball leaves the spring. So that should be H - h1 = 4.2

So that turns out

9.08 m/s

Just like the solution manual!

Woo, that was a real tough one! :smile: Thanks a bunch for all your help! I'll try more and see if I can crack 'em on my own :wink:

Mercy
 

FAQ: Conservation of mechanical energy (no friction) - doesn't work in this case?

What is conservation of mechanical energy?

Conservation of mechanical energy is a fundamental principle in physics that states that the total amount of mechanical energy (kinetic energy + potential energy) in a closed system remains constant, i.e. it is conserved. This means that energy cannot be created or destroyed, only transferred between different forms.

Why doesn't conservation of mechanical energy apply in cases with no friction?

In cases with no friction, the total mechanical energy in a system is still conserved. However, since there is no external force acting on the system to dissipate energy, the mechanical energy remains constant and does not decrease over time. This means that the system can continue to move indefinitely, which does not align with our understanding of the real world.

How does the presence of friction affect conservation of mechanical energy?

Friction is a force that opposes motion and causes energy to be converted into heat. In systems with friction, mechanical energy is not conserved as some of it is converted into other forms of energy, such as heat or sound. This means that the total mechanical energy in the system decreases over time, and the system eventually comes to a stop.

What other factors can affect the conservation of mechanical energy?

In addition to friction, other factors such as air resistance, non-conservative forces, and external work can also affect the conservation of mechanical energy. Air resistance, for example, can cause energy to be lost as heat due to the movement of an object through air. Non-conservative forces, like tension in a rope, can also transfer energy into different forms. External work, such as a person pushing or pulling on an object, can also change the amount of mechanical energy in a system.

Can conservation of mechanical energy be violated in any situation?

No, conservation of mechanical energy is a fundamental law of physics and cannot be violated. However, in certain situations, it may appear that energy is not conserved due to the limitations of our observations or measurements. For example, in a system with friction, some of the mechanical energy may be converted to heat, which is not easily detectable. Additionally, factors like air resistance and non-conservative forces can make it seem like energy is not conserved, but in reality, it is just being transferred into different forms.

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