Conservation of energy and a spring

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Homework Help Overview

The discussion revolves around a physics problem involving the conservation of energy and the behavior of a mass-spring system. The scenario describes a 1.50 kg object dropped from a height onto a vertical spring with a specified spring constant, exploring how far the spring compresses under various conditions, including the influence of gravity on the moon and the effect of air resistance.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of energy conservation principles, questioning the inclusion of kinetic energy and gravitational potential energy in their equations. Some suggest re-evaluating the definitions of variables and the setup of the energy equations, while others propose using a quadratic approach to solve for compression.

Discussion Status

Participants are actively engaging with each other's reasoning, providing insights and corrections regarding the setup of the energy equations. There is a collaborative effort to clarify misunderstandings and refine the approach to the problem, with some participants successfully deriving one of the answers through a quadratic equation.

Contextual Notes

There are indications of confusion regarding the definitions of variables and the correct application of energy conservation principles. The discussion highlights the need for careful consideration of initial conditions and assumptions in the problem setup.

Th3Proj3ct
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Homework Statement


A 1.50 kg object is held 1.20 m above a relaxed, massless vertical spring with a force constant of 320 N/m. The object is dropped onto the spring.


Homework Equations


(a) How far does the object compress the spring?
(0.381m)

(b) How far does the object compress the spring if the same experiment is performed on the moon, where g = 1.63 m/s2?
(0.143m)

(c) Repeat part (a), but now assume that a constant air-resistance force of 0.700 N acts on the object during its motion.
(0.371m)


The Attempt at a Solution


I have all the answers already (in parenthesis), but I need help on how to find them. For A I tried
I found velocity with Vf^2=Vi^2+2ad; Vf=Sqrt(2*9.8*1.2); Vf=4.8497
- K+U=K+U
- 0 + .5kx^2 = .5mv^2 + 0
- x = Sqrt(m/k)*v
- x = .332, which isn't the right answer, but I am not sure where I went wrong.
 
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Th3Proj3ct said:

Homework Statement


A 1.50 kg object is held 1.20 m above a relaxed, massless vertical spring with a force constant of 320 N/m. The object is dropped onto the spring.


Homework Equations


(a) How far does the object compress the spring?
(0.381m)

(b) How far does the object compress the spring if the same experiment is performed on the moon, where g = 1.63 m/s2?
(0.143m)

(c) Repeat part (a), but now assume that a constant air-resistance force of 0.700 N acts on the object during its motion.
(0.371m)


The Attempt at a Solution


I have all the answers already (in parenthesis), but I need help on how to find them. For A I tried
I found velocity with Vf^2=Vi^2+2ad; Vf=Sqrt(2*9.8*1.2); Vf=4.8497
- K+U=K+U
- 0 + .5kx^2 = .5mv^2 + 0
- x = Sqrt(m/k)*v
- x = .332, which isn't the right answer, but I am not sure where I went wrong.
looks like you got the final energy on the left side and the initial on the right side, and that -K is shaky, but in any case, you forgot to include the gravitational potential energy term on one side of the equation (which side and with what sign?)
 
I don't think he meant that to be a -K, I think that was just a dash.

You shouldn't have any kinetic energy in the equation, the block falls from rest, and comes to a stop once it reaches the maximum compression.
 
PhanthomJay said:
looks like you got the final energy on the left side and the initial on the right side, and that -K is shaky, but in any case, you forgot to include the gravitational potential energy term on one side of the equation (which side and with what sign?)

Maybe this would have been better:
Kef +Usf +Ugf = Kei + Usi +Ugi
0 + .5kx^2 + 0 = .5mv^2 + 0 + 17.64(mgh = 1.5(9.8)(1.2)=17.64)
But then you get 0.5kx^2=17.64+17.64
x = 4.695

hotcommodity said:
I don't think he meant that to be a -K, I think that was just a dash.

You shouldn't have any kinetic energy in the equation, the block falls from rest, and comes to a stop once it reaches the maximum compression.

Yea it was just a dash, but not using the kinetic, I get
Usf = Ugi
.5kx^2 = mgh
x = sqrt(2mgh/k)
x = 0.332 agian >_<
 
I think you may be defining "h" incorrectly. I'm working on the problem as well and I'm getting a different answer. The initial height needs to be in terms of the compression distance "x," because that's when the blocks potential energy becomes zero. You should end up with a quadratic equation.
 
Th3Proj3ct said:
Maybe this would have been better:
Kef +Usf +Ugf = Kei + Usi +Ugi
0 + .5kx^2 + 0 = .5mv^2 + 0 + 17.64(mgh = 1.5(9.8)(1.2)=17.64)
But then you get 0.5kx^2=17.64+17.64
x = 4.695
There are always several ways to attack a problem. Stick with one of them. You started by calculating the velocity as the object hit the spring. Fine. If that's your start point, then your mgh term should be mgx, where x is the compressed length of the spring.
 
Thank's all, I combined what you two said and got:
0.5kx^2=17.64+14.7x;
160x^2-14.7x-17.64 which I put into the quadratic formula and got 0.381 as one of the answers. The rest should come pretty easily now. Thanks you two for all the help, I use these forums all the time and you guys really are such a huge help.
 
I know this is an old thread, but why is 14.7 negative in your final quadratic equation?
 

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