# Motion of a mass on a spring

## Homework Statement

Suppose that a 200g mass is oscillating at the end of a spring upon a horizontal surface that is frictionless. The spring can be compressed and stretched, and has a spring constant of 300 N/m. It was originally stretched a distance of 15 cm from its equilibrium (unstretched) position prior to release.
What are the values of the (a) potential energy, (b) kinetic energy, and (c) velocity of the mass when the mass is 10 cm from the equilibrium position.

KE = 1/2mv^2
PE = mgh

## The Attempt at a Solution

I can't find the kinetic energy without finding the velocity first, but I don't know how to find the velocity in this instance. I don't know how to find the potential energy in this case either.

## Answers and Replies

Start with finding the correct equation for the PE of a spring. What you've written down is correct for gravitational PE, however your spring is sitting on a horizontal surface and in this case is completely irrelevant for the problem.

The quickest way to solve the problem is to use conservation of energy.

Thank you. So is the correct formula for PE in this case, U = 1/2 kx ^2? If that is correct, I set up the equation as U = 1/2(300)(.15)^2 and get 506.25 as my answer. Did I do that correctly?

That is the correct formula, however it looks like you typed it into your calculator wrong. The 506.25 J is wrong.

Since the mass starts out at this position (0.15m), what does this tell you about the K.E. And since you know the answer to that, what does this tell you about the total energy in the system?

Since you know the total energy in the system, and you can recalculate the PE when the mass is at 10cm away from equil. (part a), can you write down a formula for finding the KE when the mass is at 10cm away from equilibrium (part b)? And now that you know what the KE is, can you find the velocity when the mass is 10cm away from equil. (part c)?

That is the correct formula, however it looks like you typed it into your calculator wrong. The 506.25 J is wrong.

Since the mass starts out at this position (0.15m), what does this tell you about the K.E. And since you know the answer to that, what does this tell you about the total energy in the system?

Since you know the total energy in the system, and you can recalculate the PE when the mass is at 10cm away from equil. (part a), can you write down a formula for finding the KE when the mass is at 10cm away from equilibrium (part b)? And now that you know what the KE is, can you find the velocity when the mass is 10cm away from equil. (part c)?

OK, I think I fixed it and got 3.375 J. So the KE would be zero because the spring hasn't moved yet? Making the total energy of the system the 3.375 J?

I think the original question wants the PE and KE while its stretched at its original distance of 15 cm prior to release and then the velocity when the mass is at 10 cm, but I could be wrong. The way it's asked is confusing.

All of (a), (b), and (c) are asking for when the mass is at 10cm. What you have done so far are the preliminary steps that are required before you can solve the question.

Now that you've done the prelim stuff, do part (a) for the mass at 10cm, then using the answer from the prelim stuff that you just did and the answer from part (a), do part (b). To do part (c) you have to use the answer from part (b).

All of (a), (b), and (c) are asking for when the mass is at 10cm. What you have done so far are the preliminary steps that are required before you can solve the question.

Now that you've done the prelim stuff, do part (a) for the mass at 10cm, then using the answer from the prelim stuff that you just did and the answer from part (a), do part (b). To do part (c) you have to use the answer from part (b).

Oh wow, good catch. I would have messed that up. So now the PE equation becomes U = 1/2(300)(.10)^2 which equals 1.5 J. Then for the KE would I subtract the total energy of 3.375 - 1.5 to get 1.875 for my KE?

You know the answer to that question! What famous conservation law did you resort to? Why is it valid in this case?

I am also working on this question. The PE is 1.5 J or 15 J? From there, how would we find the KE? I am very confused! Too many numbers!!!!

You know the answer to that question! What famous conservation law did you resort to? Why is it valid in this case?

Haha, I often don't trust myself and like to be sure I'm doing it right. I used the conservation of energy law. I'm still kind of confused which velocity formula to use in this case and how to solve that part of this question?

If you prefer, jackboom, you can solve this problem solely with equations :).

PE is 1.5 J @ 10cm from equil.

To solve for KE, write an equation for the total energy in the system. Do you know what the total energy in the system is?

Agilic:

Well, you know the value for the KE and you wrote the formula for KE in your very first post.

So far I have
U = 1/2 kx ^2
U= ½(300)(.10)^2
1.5 J
and that is IT! I have no idea what to do next or even what I'm doing for that matter. I know KE= ½(mv^2) but i don't have the velocity yet. oyyyy

Agilic:

Well, you know the value for the KE and you wrote the formula for KE in your very first post.

Right, but I don't know how to solve for velocity in this case so that KE formula isn't helping me much right now. Was I correct in that the total energy of the system is 3.375 J and I can subtract the PE at 10cm (1.5 J) from that to solve for the KE? This would give me a KE of 1.875 J. If I am correct that that is another way to solve for KE then I still don't know how to find the velocity :/

Agilic: This is just algebra, but,

$$E_K = 1/2 mv^2 \implies v^2 = 2E_K/m \implies v = (2E_K/m)^{1/2}$$.

Do you know all of the values on the RHS of the last equation? So can you solve for velocity. And yes, you are correct about the KE. I hope you realize why you are correct.

jackboom:

Follow my advice.
Step 1: "To solve for KE, write an equation for the total energy in the system."

If this statement doesn't make sense to you, then before you try to go any further in this problem, go back and read your notes or textbook on the ideas of conservation of energy, total energy, kinetic energy, and potential energy.

is the total energy 3.375 correct? I did
Total energy = 1/2kA^2
E = 1/2 (300N)(0.15cm)^2
E=3.375 J

jack: That is correct, however why is it correct? What contributes to the total energy of the system? Write a general expression for the total energy of a system.

i have everything up to the velocity! Help!!!!

You do realize that the answer is pretty much given out in full here don't you? The answer to agilic's question provides an answer to yours. Very explicitly, I have given the formula for the velocity.

However that's not the point. If you answer the questions I've laid out for you, then you will understand the problem and be able to do it on your own. If you're uninterested in learning how to apply the concepts, then don't bother asking for more help, it won't be given.

listen, i appreciate your help but you don't have to be rude about it. i;m not a mathematician and I have no interest in physics. You have to understand how confusing reading this forum is when you are confused from the start. Formulas are being tossed around left and right so I do not know which one you are talking about. Thanks for your help but you don't have to be so rude

He's not being rude, Jack. Coto has more than answered the question(s) put forth. It's up to you, if you want to learn, to read, do, and try to understand.

Just out of curiosity - if you're not a mathematician and have no interest in physics, two immediate questions spring to mind:
1: Why, exactly, are you on these forums?
2: Why are you trying to solve this problem?

If i wasn't willing to be learning, I wouldn't be on this site and signed up. And he was being rude in my opinion. I'm on here to try and figure out this homework question. I have to answer it and if I didn't care about it, I would make the answer up. But I am confused and on here seeking help and would appreciate some patience from the people who I am seeking the help from.

It's quite clear that you don't understand physics. This is fine... that you have no interest... I really don't mind, but to give you a piece of advice: How you are applying yourself to learn this is indicative of how you apply yourself to learn anything, whether you enjoy the content or not. Even if you don't care to know about physics, the ability to solve problems will be important throughout your life, so take physics as a way to learn how to effectively solve a problem.

I have asked you multiple questions in my posts which are meant to be answered by you in your posts. This is my way of guiding you towards the solution while making sure you understand the concepts along the way.