Conservation of energy with springs

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Homework Statement


A spring of negligible mass has force constant k = 1800N/m .

You place the spring vertically with one end on the floor. You then drop a book of mass 1.20kg onto it from a height of 0.500m above the top of the spring. Find the maximum distance the spring will be compressed.

Homework Equations


Ki + Ui = Kf + Uf

The Attempt at a Solution


First I calculated the velocity of the block when it makes contact with the spring and I got v = 3.96 m/s.

Then I used conservation of energy to find the spring compression distance by setting the point where the block makes contact with the spring to be zero (y=0).
So:
Ki = Uf
(1/2)mvi2 = (1/2)kx2 - mgx
Then I substituted the values in and had to use the quadratic formula to find x which I got x = 0.10898m or x = -0.0959.
I used the negative x value because the spring is being compressed downward, but my answer is still incorrect somehow.
 

Answers and Replies

  • #2
haruspex
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(1/2)mvi2 = (1/2)kx2 - mgx
Need to be careful how you're defining the positive direction there.
I used the negative x value because the spring is being compressed downward
OK, so you are defining up as positive. What sign are you using for the value of g? Check if that agrees with the above equation.
 
  • #3
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First I calculated the velocity of the block when it makes contact with the spring and I got v = 3.96 m/s.
I find 3.13m/s for v
 
  • #4
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Need to be careful how you're defining the positive direction there.

OK, so you are defining up as positive. What sign are you using for the value of g? Check if that agrees with the above equation.
Since g is pointing down, I'm assuming I should make g negative as well as x. Why is it that when I usually do problems I'm able to keep g positive and get the right answers? Such as when throwing a ball in the air, I use Ki + Ui = Kf + Uf and I always keep g positive and still get the correct answer.
 
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I find 3.13m/s for v
Oh the velocity is 3.13 m/s. Does everything else seem correct?
 
  • #6
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Why is it that when I usually do problems I'm able to keep g positive and get the right answers? Such as when throwing a ball in the air, I use Ki + Ui = Kf + Uf and I always keep g positive and still get the correct answer.
Usually you have two points you are interested in, and in each point there is only one energy term. For example only potential energy at the top and only kinetic energy at the bottom. In that case you can equate them and the sign doesn't really give much problems. Now you have two energy terms at the bottom, so you have to equate the kinetic energy at the impact with the spring to the sum of the two energy terms when the spring is maximally stretched. In this case you have to wacht whether you add them or subtract them.

Btw, the problem is the calculation of v. Not to mention that you didn't even needed that per se as you could have used the potential energy at the time of dropping the book as well.
 
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  • #7
haruspex
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Since g is pointing down, I'm assuming I should make g negative as well as x.
There are two ways people handle this. You can think of 'g' as representing a magnitude, so you write g in the equation as an acceleration of -g (up being positive), or you can treat g as representing the acceleration due to gravity in the upward positive model. In the second (which I prefer), g appears in the equation as though it is upwards, but its numerical value will be negative.
By the way, you didn't need to work out the velocity. Just consider total change in PE.
 
  • #8
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Usually you have two points you are interested in, and in each point there is only one energy term. For example only potential energy at the top and only kinetic energy at the bottom. In that case you can equate them and the sign doesn't really give much problems. Now you have two energy terms at the bottom, so you have to equate the kinetic energy at the impact with the spring to the sum of the two energy terms when the spring is maximally stretched. In this case you have to wacht wether you add them or subtract them.

Btw, the problem is the calculation of v. Not to mention that you didn't even needed that per se as you could have used the potential energy at the time of dropping the book as well.
If it is just two points such as in the example with the baseball, finding the velocity will create be a problem if you make g negative though. Because at the top it will be -mgh which will equal the initial kinetic energy, (1/2)mv^2 and if you had to solve for velocity, you would have to take the square root of a negative value.
 
  • #9
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Yes, but I was responding on why keeping g positive wasn't giving a lot of problems. By the way the inability to take a square root of a negative value helps as well, because most people will recognize that you can't do that and adjust accordingly
 
  • #10
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I'm still very confused. How is it possible that I can randomly change the sign of gravity just because it isn't working in my equation. By changing gravity from negative to positive, I changed the direction of gravity to up instead of down. If I am able to do this then I should be able to keep it positive for my original problem. Originally I always kept acceleration due to gravity negative when I'm defining up as positive, but once I got to conservation of energy I wasn't able to anymore which is why I started using g as positive.
 
  • #11
haruspex
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How is it possible that I can randomly change the sign of gravity just because it isn't working in my equation.
You can't. Did you read my post #7? Can you post a specific example of the sign of g not mattering?
 
  • #12
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You can't. Did you read my post #7? Can you post a specific example of the sign of g not mattering?
If a baseball is thrown straight up and reaches the maximum height, what is the initial velocity.
Ui + Ki = Uf + Kf
Ui = Kf = 0.
Then Ki = Uf.
And if you set g to be negative in Uf, you have to take a square root of a negative.
 
  • #13
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How is it possible that I can randomly change the sign of gravity just because it isn't working in my equation.
You can't change randomly the sign of gravity. You can however define upwards to be positive or downwards to be positive. g is ALWAYS downwards, so in the first case the acceleration is negative, in the second the acceleration is positive. As soon as you define something and you are consistent in your approach throughout the entire problem it shouldn't give problems

By changing gravity from negative to positive, I changed the direction of gravity to up instead of down.
No you have defined downwards as the positive direction.

If a baseball is thrown straight up and reaches the maximum height, what is the initial velocity.
Ui + Ki = Uf + Kf
Ui = Kf = 0.
Then Ki = Uf.
And if you set g to be negative in Uf, you have to take a square root of a negative.
I missed to mention a part of your problem with the signs.
The potential energy is actually the weight times the height and the weight of something is the mass times the magnitude of g and is always positive. So when looking at potential energies, they can be negative only if your height is negative. And this of course depends on what height you take to be 0.
If you look at forces for example then that is different because you are looking at g not the magnitude of g

You do have to watch out if x is positive or negative
 
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  • #14
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You can't change randomly the sign of gravity. You can however define upwards to be positive or downwards to be positive. g is ALWAYS downwards, so in the first case the acceleration is negative, in the second the acceleration is positive. As soon as you define something and you are consistent in your approach throughout the entire problem it shouldn't give problems



No you have defined downwards as the positive direction.



I missed to mention a part of your problem with the signs.
The potential energy is actually the weight times the height and the weight of something is the mass times the magnitude of g and is always positive. So when looking at potential energies, they can be negative only if your height is negative. And this of course depends on what direction you call positive.
If you look at forces for example then that is different because you are looking at g not the magnitude of g

You do have to watch out if x is positive or negative
That makes much more sense now. Thank you so much for the explanation!
 
  • #15
haruspex
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Then Ki = Uf.
And if you set g to be negative in Uf, you have to take a square root of a negative.
U(r) = -mMG/r.
ΔU(x) = U(R+x)-U(R) = mMG(1/R - 1/(R+x)) ~= mMGx/R2 > 0.
g = -MG/R2
ΔU(x) = -mgx.
x > 0 & g < 0 gives ΔU > 0.
 
  • #16
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U(r) = -mMG/r.
ΔU(x) = U(R+x)-U(R) = mMG(1/R - 1/(R+x)) ~= mMGx/R2 > 0.
g = -MG/R2
ΔU(x) = -mgx.
x > 0 & g < 0 gives ΔU > 0.
After reading over everything again I'm still confused as to why the sign of gravity is negative in my original problems simply because it is now under the reference point. I feel as though only the height will be negative now because it is below zero. Why would I need gravity specifically to be negative in this equation and not in the baseball example?
 
  • #17
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Also, now that I'm under the reference point, it seems as though gravitational potential energy should be negative as it is in my equation in the first post, but I only got the correct answer if I made it positive.
 
  • #18
haruspex
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Why would I need gravity specifically to be negative in this equation and not in the baseball example?
It is negative in the baseball example:
x > 0 & g < 0 gives ΔU > 0.
Also, now that I'm under the reference point, it seems as though gravitational potential energy should be negative as it is in my equation in the first post, but I only got the correct answer if I made it positive.
The change in PE, in the question at start of thread, is negative. I can't see any post where you show all your working and get the right answer.
 
  • #19
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It is negative in the baseball example:


The change in PE, in the question at start of thread, is negative. I can't see any post where you show all your working and get the right answer.
So I found the way to get the correct answer is by using (1/2)mvi2 = (1/2)kx2 - mgx.
After solving for x using the quadratic equation, there is a negative and positive x value. The correct answer is the positive x value which is 0.088 m. Is there a reason the positive answer is the correct one? Because if the spring is being compressed below the reference point, shouldn't I take the negative x value?
 
  • #20
haruspex
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So I found the way to get the correct answer is by using (1/2)mvi2 = (1/2)kx2 - mgx.
After solving for x using the quadratic equation, there is a negative and positive x value. The correct answer is the positive x value which is 0.088 m. Is there a reason the positive answer is the correct one? Because if the spring is being compressed below the reference point, shouldn't I take the negative x value?
There are two solutions for x because the system will oscillate. At both extremes, the KE will be zero, and all the energy will be as a combo of gravitational PE and spring PE. Your equation encompasses both cases.
Which one is the solution required depends on the value you are putting in for g. If g = -9.8 m/s2, then your equation has x as positive up, and you should take the negative value of x.
 

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