# Conservation of energy -- Using a spring to launch a ball up an incline

• tootired
In summary, you are trying to calculate the speed of a .164 kg ball when it leaves a compressed spring. You convert N/cm to N/m and get a value of 1.09 m/s.
tootired
Homework Statement
Launcher has a spring force constant of 1.52N/cm. The surface is at an incline of 8.3 degrees.
The Spring is initially compressed 3.75 cm. Find the launching speed of the .164 kg ball when the plunger is released. A due to G is 9.8m/s^2. Friction and the plunger are negligible
Relevant Equations
Ei=Ef
Ei = 1/2 K (x)^ 2
K = .0152N/m
x = .0375 m
Ei = 1.06x10^-5

Ef= 1/2mv2 + mgh
m = .164kg, v is unknown, h is .0375sin(8.3)=.00541, Ef set equal to Ei
1.06x10^-5=1/2(.164kg)(v^2)+ (.164kg)(9.8)(.00541)
v = .3254m/s

I have gotten this answer multiple times but it is not correct. I am going a bit crazy, someone please give me clue here

tootired said:
Homework Statement:: Launcher has a spring force constant of 1.52N/cm. The surface is at an incline of 8.3 degrees.
The Spring is initially compressed 3.75 cm. Find the launching speed of the .164 kg ball when the plunger is released. A due to G is 9.8m/s^2. Friction and the plunger are negligible
Relevant Equations:: Ei=Ef

Ei = 1/2 K (x)^ 2
K = .0152N/m
x = .0375 m
Ei = 1.06x10^-5

Ef= 1/2mv2 + mgh
m = .164kg, v is unknown, h is .0375sin(8.3)=.00541, Ef set equal to Ei
1.06x10^-5=1/2(.164kg)(v^2)+ (.164kg)(9.8)(.00541)
v = .3254m/s

I have gotten this answer multiple times but it is not correct. I am going a bit crazy, someone please give me clue here
Hi @tootired and welcome to PF.

You are given that k = 1.52 N/cm. This means that when you stretch the spring by 1 cm it pulls back with a force of 1.52 Newtons.

You say that k = .0152 N/m. This means that when you stretch the spring 100 times as much as before, it pulls back with a force that is 100 times less. Huh?

You need some rest because it seems you are too tired.

tootired and Lnewqban
kuruman said:
Hi @tootired and welcome to PF.

You are given that k = 1.52 N/cm. This means that when you stretch the spring by 1 cm it pulls back with a force of 1.52 Newtons.

You say that k = .0152 N/m. This means that when you stretch the spring 100 times as much as before, it pulls back with a force that is 100 times less. Huh?

You need some rest because it seems you are too tired.
I would agree on the rest, but not an option lately. If I do not convert N/cm to N/m the answer is still not correct. I'm all out of attempts on this one. Thank you for the insight

Welcome!

How is the "incline of 8.3 degrees" relevant in this problem?
Uphill, I assume.
Could you show us a diagram?

Please, read this guide about working with symbolic variables as far as possible in the calculation process:

It seems that the calculated value of Ei is incorrect, but units are not shown.
Energy of the compressed spring is 10.69 N-cm.
Why don't you work with N and cm and convert at the end to m/s?

Last edited:
tootired said:
If I do not convert N/cm to N/m the answer is still not correct.
Not sure what you mean by that. You do need to convert at some point, but to do so correctly. You should have noticed that the expression you got for ##v^2## was negative.
In case you are still getting the conversion wrong, here's the safest way to do it. There are 100 cm per metre, so ##1.52N/cm\times 100cm/m=152N/m##. This should yield an answer about 1.09m/s.
Lnewqban said:
How is the "incline of 8.3 degrees" relevant in this problem?
Because as post #1 working shows, some energy is used in raising the ball to the point where it leaves the spring. The angle is needed to find the height gain.

Lnewqban
The ball will lose contact at a point where its acceleration is zero. That point is the equilibrium position of the inclined spring-mass system which occurs where the spring is compressed by ##x_{\text{eq}}=mg\sin\!\theta/k##. If we assume that the given ##x_0=3.75## cm is the compression from the relaxed length of the spring, the energy conservation equation in the form ##\Delta U+\Delta K=0## would be $$\frac{1}{2}k\left(x_{\text{eq}}^2-x_0^2 \right)+mg\sin\!\theta(x_{\text{eq}}-x_0)+\frac{1}{2}m\left(v^2-0\right)=0.$$In this case, the incline angle is small and ##x_{\text{eq}}=0.15## cm or 4% of ##x_0##. This may or may not be sufficient to push the approximate solution with ##x_{\text{eq}}=0## beyond the tolerance of the scoring algorithm.

Lnewqban
kuruman said:
The ball will lose contact at a point where its acceleration is zero.
No, the spring, being ideal, will expand as fast as necessary to maintain contact until it reaches its relaxed length. At some point before that the force it exerts on the ball will be insufficient to overcome gravity, so the ball reaches its maximum speed there, but that does not alter the correctness of the energy calculation for the speed at the moment contact is lost.

haruspex said:
No, the spring, being ideal, will expand as fast as necessary to maintain contact until it reaches its relaxed length.
Indeed. I was thinking of the other problem with two separate masses one of which is attached to the spring and the other buting against the first mass.

## 1. How does a spring launch a ball up an incline?

When a spring is compressed, it stores potential energy. This potential energy is then converted into kinetic energy when the spring is released, causing the ball to move upwards along the incline.

## 2. What is the conservation of energy principle?

The conservation of energy principle states that energy cannot be created or destroyed, only transferred or converted from one form to another. This means that the total energy of a closed system remains constant.

## 3. How does the conservation of energy apply to using a spring to launch a ball?

In this scenario, the potential energy stored in the compressed spring is converted into kinetic energy as the spring is released. The kinetic energy then propels the ball upwards along the incline. As the ball moves, some of its kinetic energy is converted into gravitational potential energy due to its increasing height. However, the total energy of the system (spring + ball) remains constant.

## 4. Does the mass of the ball affect the conservation of energy in this experiment?

Yes, the mass of the ball does affect the conservation of energy. The heavier the ball, the more potential energy it will have as it moves up the incline. However, the total energy of the system will still remain constant.

## 5. Are there any external factors that could affect the conservation of energy in this experiment?

Yes, there are external factors that could affect the conservation of energy. Friction between the ball and the incline, as well as air resistance, can cause some of the energy to be lost as heat. This means that the ball may not reach the same height on subsequent launches, as some of the energy has been dissipated. Additionally, the accuracy of the spring and the incline's angle can also affect the conservation of energy in the experiment.

• Introductory Physics Homework Help
Replies
10
Views
772
• Introductory Physics Homework Help
Replies
3
Views
543
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
5
Views
963
• Introductory Physics Homework Help
Replies
6
Views
520
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
17
Views
546
• Introductory Physics Homework Help
Replies
2
Views
718
• Introductory Physics Homework Help
Replies
5
Views
1K