# SOLVED: Height Cheese Rises After Spring Released

• clope023
In summary, the piece of cheese with a mass of 1.35kg, placed on a vertical spring of negligible mass and a force constant k = 1600N/m that is compressed by a distance of 12.3cm, will rise to a height of approximately 0.92m when the spring is released. The initial position is measured from the lowest point of the compressed spring, with the height of the cheese being measured from the same reference point.

#### clope023

[SOLVED] Released Spring

## Homework Statement

A piece of cheese with a mass of 1.35kg is placed on a vertical spring of negligible mass and a force constant k = 1600N/m that is compressed by a distance of 12.3cm .

When the spring is released, how high does the cheese rise from this initial position? (The cheese and the spring are not attached.)

## Homework Equations

U1 = Ugrav1 + Uel1

U2 = Ugrav2

v1 = 0

v2 = 0

K1 = 0

K2 = 0

K1 + U1 = K2 + U2

--> Ugrav1 + Uel1 = Ugrav2

y1 = -.123m

## The Attempt at a Solution

(1.35kg)(9.81m/s^2)(-.123m) + 1/2k(-.123m)^2 = (1.35kg)(9.81m/s^2)h

h = ((1.35kg)(9.81m/s^2)(-.123m) + 1/2(1600N/m)(-.123m)^2))/((1.35kg)(9.81m/s^s))

h = .80m

it's incorrect and I'm not sure what I'm doing wrong, at the bottom position where the spring is compressed the initial position should be less than 0 correct?

there's no velocity at the inital position and at the highest point there is no velocity or potential energy, only gravity should be acting on it correct? any help is greatly appreciated.

clope023 said:

## Homework Statement

A piece of cheese with a mass of 1.35kg is placed on a vertical spring of negligible mass and a force constant k = 1600N/m that is compressed by a distance of 12.3cm .

When the spring is released, how high does the cheese rise from this initial position? (The cheese and the spring are not attached.)
The phrase "this initial position" is ambiguous. I would assume they mean from the lowest position of the compressed spring, not the unstretched position.

Doc Al said:
The phrase "this initial position" is ambiguous. I would assume they mean from the lowest position of the compressed spring, not the unstretched position.

wouldn't the lowest position be the distance it was compressed?

what I'm doing is I'm taking the positions to be (-.123m, h), h being the variable I'm trying to find.

bump, anyone? I don't see how the initial position can be anything else.

clope023 said:
wouldn't the lowest position be the distance it was compressed?

what I'm doing is I'm taking the positions to be (-.123m, h), h being the variable I'm trying to find.
You have defined your height (h) from the unstretched position, not the lowest position.

Doc Al said:
You have defined your height (h) from the unstretched position, not the lowest position.

but wouldn't the unstretched position be when the cheese is not on the spring? I'm trying to grasp how the cheese's mass compressing it a distance of .123m isn't the lowest position? would it then be .246m?

do I have to find integrals first to find the lowest position and then from that position to the height using the spring force?

edit: sorry if I'm being difficult, just not getting the positions very well here I guess, I do have the right formulas I think right?

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clope023 said:
but wouldn't the unstretched position be when the cheese is not on the spring? I'm trying to grasp how the cheese's mass compressing it a distance of .123m isn't the lowest position?
It is the lowest position. But you are measuring heights from the unstretched position, not the lowest position: You call the initial height = -.123 m (not zero).

Doc Al said:
It is the lowest position. But you are measuring heights from the unstretched position, not the lowest position: You call the initial height = -.123 m (not zero).

but that's what I did; I used y1 = -.123m and plugged it into the initial gravitational and elastic energies, I didn't use 0.

clope023 said:
but that's what I did; I used y1 = -.123m and plugged it into the initial gravitational and elastic energies, I didn't use 0.
Answer this question: If something starts at y = - 5m and rises to the point y = 10m, how far does it move relative to the starting point?

Doc Al said:
Answer this question: If something starts at y = - 5m and rises to the point y = 10m, how far does it move relative to the starting point?

15 meters

Good. Now apply that same reasoning to your problem, measuring from the starting point (the lowest point).

Doc Al said:
Good. Now apply that same reasoning to your problem, measuring from the starting point (the lowest point).

the original answer I got, .80m

then add to that the distance from y0 to h, and solve for h.

though y0 = 0 so I'm not sure what kind of equation I'm supposed to use since gravitational potentional energy will go to zero there.

clope023 said:

the original answer I got, .80m

then add to that the distance from y0 to h, and solve for h.

though y0 = 0 so I'm not sure what kind of equation I'm supposed to use since gravitational potentional energy will go to zero there.
In your original conservation of energy equation, there are gravitational PE terms on both sides which involve heights. Those heights must be measured from the same reference point (otherwise it makes no sense). On the left, you chose to measure height from the unstretched position, thus y1 = -.123 m. Thus you must interpret your answer for h from the same reference point.

Then you can translate your value for h into what the question asked: Height from the starting point (which we assume is the lowest point).

You can redo your equation, using y1 = 0. (Nothing wrong with gravitational PE = 0!) Or you can just properly interpret your existing value for h in terms of what the question asked. (No need to do anything over.)

Doc Al said:
In your original conservation of energy equation, there are gravitational PE terms on both sides which involve heights. Those heights must be measured from the same reference point (otherwise it makes no sense). On the left, you chose to measure height from the unstretched position, thus y1 = -.123 m. Thus you must interpret your answer for h from the same reference point.

Then you can translate your value for h into what the question asked: Height from the starting point (which we assume is the lowest point).

You can redo your equation, using y1 = 0. (Nothing wrong with gravitational PE = 0!) Or you can just properly interpret your existing value for h in terms of what the question asked. (No need to do anything over.)

I'm sorry but I don't understand what you're saying, or rather I don't get how to write it mathematically, are you saying I should be use.

mg(-.123) + 1/2k(-.123)^2 = mg(-.123)?

or mgh + 1/2kh^2 = mg(-.123)?

also using y1 = 0, how would I get an answer than, since:

mg0 + 1/2k0 = mgh would just give me 0?

and how is -.123m the 'unstretched' position?

sorry again but I'm just not understanding what you're saying.

okay, I just go the answer, there's no potential energy at positon 0 or -.123m, but there is elastic energy at position -.123m, and there's gravitational energy at positon h, so what I did was:

1/2ky^2 = mgh

h = 1/2ky^2/mg

h = (1/2(1600)(-.123m)^2))/(1.35*9.81) = .91m, it was the exact answer,

it also turns out that my initial answer of .79 + .123m would have also given me .91m, thanks Doc Al for the help (and putting up with my nonunderstanding )

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I knew you'd figure it out, sooner or later.

## What is the experiment about?

The experiment is about observing the relationship between the height of cheese and the release of a spring.

## What were the results of the experiment?

The results showed that the height of the cheese increased after the spring was released.

## What is the significance of the experiment?

This experiment demonstrates the principle of potential energy being converted into kinetic energy, as the spring releases and causes the cheese to rise.

## What factors may have influenced the results?

The results may have been influenced by the type and strength of the spring used, the weight and type of cheese, and the height from which the spring was released.

## How can this experiment be applied in real life?

This experiment can be applied in real life to understand the principles of potential and kinetic energy, which have practical applications in fields such as engineering and physics.

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