Rock on Vertical Spring: Calculating Maximum Height and Potential Energy

In summary, the stone is pushed down an additional 30.0 cm and released. The change in gravitational potential energy of the stone-Earth system is 76.7 joules.
  • #1
mbrmbrg
496
2
An 8.00 kg stone rests on a spring. The spring is compressed 7.0 cm by the stone.

(a) What is the spring constant?
11.2 N/cm
(b) The stone is pushed down an additional 30.0 cm and released. What is the elastic potential energy of the compressed spring just before that release?
76.7 J
(c) What is the change in the gravitational potential energy of the stone-Earth system when the stone moves from the release point to its maximum height?
76.7 J
(d) What is that maximum height, measured from the release point?

a, b, and c are correct. part d just confuses me.
I said:
[tex]mgh_f - mgh_i = \Delta U_G=76.7J[/tex] (as per part c)

so [tex]mgh_f = 76.7J + mg(-0.37m)[/tex]

so [tex]h_f=\frac{76.7-(8)(9.81)(0.37)}{(8)(9.81)}=0.608[/tex]

Lovely WebAssign puts next to this a lovely red x. Somehow, that makes me wonder if perhaps my answer to part (d) is wrong...
 
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  • #2
as you said
(a) 11.2 N/cm or 1120 N/m correct
(b) 76.7 J correct
(c) 76.7 J correct
(d) I get something different
It says "measured from the release point", NOT from equilibrium point as you assumed. So you simply made a silly mistake, nothing big.
 
  • #3
mbrmbrg said:
An 8.00 kg stone rests on a spring. The spring is compressed 7.0 cm by the stone.

(a) What is the spring constant?
11.2 N/cm
(b) The stone is pushed down an additional 30.0 cm and released. What is the elastic potential energy of the compressed spring just before that release?
76.7 J
(c) What is the change in the gravitational potential energy of the stone-Earth system when the stone moves from the release point to its maximum height?
76.7 J
(d) What is that maximum height, measured from the release point?

a, b, and c are correct. part d just confuses me.
I said:
[tex]mgh_f - mgh_i = \Delta U_G=76.7J[/tex] (as per part c)

so [tex]mgh_f = 76.7J + mg(-0.37m)[/tex]

so [tex]h_f=\frac{76.7-(8)(9.81)(0.37)}{(8)(9.81)}=0.608[/tex]

Lovely WebAssign puts next to this a lovely red x. Somehow, that makes me wonder if perhaps my answer to part (d) is wrong...
The gravitational potential energy change of 76.7 joules is in reference to the release point where the initial GPE is established as 0. So mgh_i = 0.
h = 76.7/mg as measured from the release point
 
  • #4
But how do I calculate the release point? I have no idea how far the spring can stretch!
 
  • #5
Your equation implies that Ug = 0 @ 37 cm above the release point. Given the conditions of the problem, that is the equilibrium point of the spring. So the final height you get is the height above the equilibrium point, that's not what you want.

If you want the height above the release point (the point where the spring is initial set into motion, after being compressed another 30 cm), then set Ug = 0 at that point. Then mgh_i = 0.

This is basically what PhantomJay is telling you, and I think he is definitely correct.

Dorothy
 
  • #6
mbrmbrg said:
But how do I calculate the release point? I have no idea how far the spring can stretch!
The spring is not going to stretch. It is assumed massless, so it has no momentum once the stone is gone. The stone is not connected to the spring. The release point is where the stone is when the spring is at the maximum stated compression.
 
  • #7
mbrmbrg said:
But how do I calculate the release point? I have no idea how far the spring can stretch!
By release point, the problem means the point at which you let go of the rock. So right where the stone is before it gets launched, when it is still stationary.
 
Last edited:
  • #8
Oh! I assumed that the release point was the point where the rock no longer touched the spring; ie where the spring was stretched to its maximum (so the spring would start moving toward Earth while the rock continued moving upward).

Thank you, all!
 

1. What is a "Rock on Vertical Spring"?

A "Rock on Vertical Spring" is a simple physics demonstration that involves a small rock attached to a vertical spring. When the rock is pulled down and released, it will oscillate up and down on the spring until friction and air resistance cause it to eventually come to a stop.

2. What is the purpose of this experiment?

The purpose of this experiment is to demonstrate the concept of simple harmonic motion, which occurs when a system oscillates back and forth at a constant frequency. It also allows for the study of factors that affect the motion, such as the mass of the object, the spring constant, and the amplitude of the oscillation.

3. What factors affect the motion of the rock on the vertical spring?

The motion of the rock on the vertical spring can be affected by several factors, including the mass of the rock, the spring constant of the spring, the amplitude of the oscillation, and external factors such as air resistance and friction.

4. How can this experiment be used to study simple harmonic motion?

This experiment can be used to study simple harmonic motion by changing the variables and observing how they affect the motion of the rock. For example, changing the mass of the rock or the spring constant of the spring can affect the frequency and amplitude of the oscillation, allowing for a better understanding of the principles of simple harmonic motion.

5. What are some real-life applications of simple harmonic motion?

Simple harmonic motion can be observed in many real-life situations, such as the motion of a pendulum, the vibrations of a guitar string, or the movement of a spring in a car's suspension system. It is also used in the design of various mechanical systems, such as clocks, watches, and musical instruments.

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