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Elastic Potential energy problem (simple)

  • #1
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Homework Statement


A 1.20 kg piece of cheese is placed on a vertical spring of negligible mass and force constant k = 1800 N/m that is compressed 15 cm. When the spring is released, how high does the cheese rise from this initial position? (The cheese and the spring are not attached).

I looked up the solution to cramster, and the person did something bewildering (bewildering to me at least).

He set [itex] u=mgh=1/2kx^2[/itex] then solved for h and got [itex]s = 1.72m[/itex], but this doesn't make sense to me because he didn't take the acceleration of gravity in effect? I believe mine is more correct but I don't know too much in elastic potential energy.


Homework Equations



[tex]v^2=V_o^2+2as[/tex]
[itex]K)1+u_g+u_{el}=k_2+U_{g2}+U_{el2}[/itex]
I set the relaxed state of the spring to be 0.

The Attempt at a Solution



[tex]mgh_1+1/2kx^22=1/2mv^2+mgh_2+1/2kx^2[/tex]
[tex]mgh_2[/tex]and [tex]1/2kx^2_2[/tex] go to 0
[tex]1.2(9.8)(-.15) + 1/2(1800)(-.15)^2 = 1/2 (1.2)v^2[/tex]
Solve for v accordingly
[tex]v=5.55 m/s[/tex]

Then I use a simple kinematics equation to see how high it will go in the air.

[itex]v_o = 5.55m/s[/itex] [tex]a=9.8m/s^2[/tex] [tex]v_f = 0[/tex] [tex]s=?[/tex] [tex]v^2= v_o^2 + 2as[/tex] When we solve for s we get [itex]s=1.57m[/itex]
 
Last edited:

Answers and Replies

  • #2
Doc Al
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He set [itex] u=mgh=1/2kx^2[/itex] then solved for h and got [itex]s = 1.72m[/itex], but this doesn't make sense to me because he didn't take the acceleration of gravity in effect?
The effect of gravity is automatically taken care of by the gravitational PE term. Since energy is conserved, you can compare the initial position (where the spring is compressed) to the final position (where the cheese is at its highest point) without worrying about the details in between.

I believe mine is more correct but I don't know too much in elastic potential energy.


Homework Equations



[tex]v^2=V_o^2+2as[/tex]
[itex]K)1+u_g+u_{el}=k_2+U_{g2}+U_{el2}[/itex]
I set the relaxed state of the spring to be 0.

The Attempt at a Solution



[tex]mgh_1+1/2kx^22=1/2mv^2+mgh_2+1/2kx^2[/tex]
[tex]mgh_2[/tex]and [tex]1/2kx^2_2[/tex] go to 0
[tex]1.2(9.8)(-.15) + 1/2(1800)(-.15)^2 = 1/2 (1.2)v^2[/tex]
Solve for v accordingly
[tex]v=5.55 m/s[/tex]

Then I use a simple kinematics equation to see how high it will go in the air.

[itex]v_o = 5.55m/s[/itex] [tex]a=9.8m/s^2[/tex] [tex]v_f = 0[/tex] [tex]s=?[/tex] [tex]v^2= v_o^2 + 2as[/tex] When we solve for s we get [itex]s=1.57m[/itex]
Realize that the problem asked for how high the cheese rose from the initial position (when the spring is compressed). So you need to add the 15 cm to your answer.

Your method is fine, but it involves a bit of unnecessary work. And you get the same answer, of course.
 
  • #3
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The effect of gravity is automatically taken care of by the gravitational PE term. Since energy is conserved, you can compare the initial position (where the spring is compressed) to the final position (where the cheese is at its highest point) without worrying about the details in between.
I know the gravity is taken care of in gravitational PE but how does he set mgh = 1/2kx^2 ?

Realize that the problem asked for how high the cheese rose from the initial position (when the spring is compressed). So you need to add the 15 cm to your answer.
Oh okay. =d

Your method is fine, but it involves a bit of unnecessary work. And you get the same answer, of course.
Yes for some reason I usually end up doing it the longer way in Elastic PE problems haha. There is something I don't understand that I'm still trying to pinpoint what it is.
 
  • #4
Doc Al
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I know the gravity is taken care of in gravitational PE but how does he set mgh = 1/2kx^2 ?
He's just setting final energy (purely gravitational PE) to initial energy (purely elastic PE). Note that he measures gravitational PE from the initial position.
 
  • #5
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He's just setting final energy (purely gravitational PE) to initial energy (purely elastic PE). Note that he measures gravitational PE from the initial position.
To me it doesn't make sense not to include kinetic energy, after all kinetic energy is involved is it not?

E= K + U is constant
which is;

[tex]E = k + u_g + u_l[/tex]
 
  • #6
gneill
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To me it doesn't make sense not to include kinetic energy, after all kinetic energy is involved is it not?

E= K + U is constant
which is;

[tex]E = k + u_g + u_l[/tex]
At the start the cheese is not moving. So KE = 0. At the highest elevation, the cheese is not moving, so KE = 0. So KE drops out of consideration if you are comparing the starting and ending conditions.

If you wanted to consider some intermediate point in the trajectory, then you would have to figure in the KE for that instant.
 
  • #7
Doc Al
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To me it doesn't make sense not to include kinetic energy, after all kinetic energy is involved is it not?

E= K + U is constant
which is;

[tex]E = k + u_g + u_l[/tex]
It's not that you aren't including it, it's just that it's zero at those points of interest.

The total mechanical energy at every point in the motion is the sum of all three kinds of energy. For this particular problem, there's no need to explore the details of the motion at intermediate points as it leaves the spring and rises up. But you could, at any point, figure out just how fast it's moving if you wanted to.

The beauty of energy conservation is that it often allows you to solve for things without having to worry about the details of the motion.
 
  • #8
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Oh okay.. So the way I thought about it was to break it up into two pieces.

1) From starting position to position as spring is non-compressed (0 elastic PE).
2) Then from position of non-compressed spring to the final velocity in the air (maximum height) where it is 0.

But I guess, as gneill mentioned, that you can just skip the intermediate step and calculate from starting position to final height.. You can't do this for a lot of problems in physics.. I guess this is an exception but I can't completely internalize when to use it. I'll just perform the extra steps for subsequent problems just in case. Better safe then sorry. :)

Thanks for the help gneill and Doc AL.
 
  • #9
Doc Al
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But I guess, as gneill mentioned, that you can just skip the intermediate step and calculate from starting position to final height.. You can't do this for a lot of problems in physics.. I guess this is an exception but I can't completely internalize when to use it. I'll just perform the extra steps for subsequent problems just in case. Better safe then sorry. :)
The thing to ask yourself is whether energy is conserved or not. Whenever mechanical energy is conserved, you can skip the intermediate steps.
 

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