Solving Spring Questions: 4.6 cm and 6.7 cm

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The discussion focuses on solving two spring stretch problems involving different masses and spring constants. The first scenario involves a 4.72-kg mass causing a spring with a constant of 20 N/cm to stretch incorrectly calculated as 2.3 cm, while the correct answer is 4.6 cm. The second scenario correctly calculates a 4-kg mass stretching a spring with a constant of 5.9 N/cm to 6.7 cm. The key difference between the two situations is the initial conditions: one mass is dropped (with kinetic energy) while the other is lowered slowly (static equilibrium). Understanding the distinction between these conditions is crucial for applying the correct physics principles.
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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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