Solving Spring Questions: 4.6 cm and 6.7 cm

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Homework Help Overview

The discussion revolves around two problems involving the stretching of springs when masses are attached. The first problem involves a 4.72-kg mass and a spring constant of 20 N/cm, while the second involves a 4-kg mass with a spring constant of 5.9 N/cm. Participants explore the differences in outcomes based on how the masses are released.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of force equations and the distinction between letting a mass fall versus slowly lowering it. Questions arise regarding the implications of kinetic energy in the context of static equilibrium.

Discussion Status

Some participants have provided insights into the differences between the two scenarios, suggesting that one involves kinetic energy while the other does not. There is an ongoing exploration of how these concepts fit into the equations being used.

Contextual Notes

Participants are navigating the complexities of energy versus force in the context of spring mechanics, with specific attention to the assumptions made in each scenario. The original poster expresses confusion about the application of concepts across different problem setups.

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Homework Statement


1. An unstretched spring with spring constant 20 N/cm is suspended from the ceiling. A 4.72-kg mass is attached to the spring and let fall. To the nearest tenth of a centimeter, how far does it stretch the spring?

2. An unstretched spring with spring constant 5.9 N/cm is suspended from the ceiling. A 4-kg mass is attached to the spring and slowly let down. To the nearest tenth of a centimeter, how far does it stretch the spring?

Homework Equations



F=-kx
F=ma

The Attempt at a Solution


1.
ma=kx
(4.72 kg)(9.81 m/s^2)=(2000 N/m)(x)
x=2.3 cm
Wrong. The answer is 4.6 cm.

2.
ma=kx
(4 kg)(9.81 m/s^2)=(590 N/m)(x)
x=6.7 cm
Correct. Even though I used the same steps, this one is correct and the other is wrong. Am I missing a crucial difference between "let fall" and "slowly let down"?

Thanks in advance!
 
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Am I missing a crucial difference between "let fall" and "slowly let down"?
yes you are. All bodies in motion have kinetic energy equal to mv2/2
Because one is let down (V=0), and the other is dropped (V=/=0)

In other words, you assumed both were static equilibrium problems, when in reality only the second one is.

Hope that helps :)
 
Thank you, that makes sense.

How does this fit into the equations though, since I am using forces and this relates to energy?
 
Simply remember the definition of Work, W=F*d or W=integrate(F,d,d1,d2)
If you're in calc based physics, you should be able to derive the equation, if not, then find the equation for PE of a spring on your equation sheet.
 
Oh, I see.

Just for future reference, this is the solution:

Potential energy due to gravity is mgh; h in this case is x.

mgx=1/2kx^2
mg=1/2kx
(4.72 kg)(9.81 m/s^2)=1/2(2000 N/m)(x)
x=4.6 cm

Thanks again for the help!
 

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