Solving Spring Questions: 4.6 cm and 6.7 cm

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In summary, two scenarios are given where an unstretched spring with different spring constants is attached to a mass and suspended from the ceiling. The first scenario involves the mass being let fall, while the second scenario involves it being slowly let down. The solution for the first scenario is incorrect because it assumes static equilibrium, while the second scenario is correct because it takes into account the initial kinetic energy of the mass. To correctly solve for the distance the spring stretches, potential energy due to gravity and the equation for potential energy of a spring are used. The correct answer for the first scenario is 4.6 cm.
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Hypercubes
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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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  • #2
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 :)
 
  • #3
Thank you, that makes sense.

How does this fit into the equations though, since I am using forces and this relates to energy?
 
  • #4
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.
 
  • #5
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!
 

FAQ: Solving Spring Questions: 4.6 cm and 6.7 cm

How do you calculate the spring constant of a spring?

The spring constant of a spring can be calculated by dividing the force applied to the spring by the displacement of the spring from its original position. It is represented by the letter k and has the unit N/m in the SI system.

What is the difference between the initial length and the final length of a spring?

The initial length of a spring is the length of the spring when no external force is applied to it. The final length, on the other hand, is the length of the spring when an external force is applied to it, causing it to stretch or compress.

How do you determine the displacement of a spring?

The displacement of a spring can be determined by measuring the difference between the final length and the initial length of the spring. This can be calculated by subtracting the initial length from the final length, or by measuring the change in length using a ruler or other measuring device.

What is Hooke's law and how does it relate to springs?

Hooke's law states that the force applied to a spring is directly proportional to the displacement of the spring from its original position. This means that the more a spring is stretched or compressed, the greater the force applied to it. This law is used to calculate the spring constant of a spring.

Can the spring constant of a spring change?

Yes, the spring constant of a spring can change depending on the material it is made of, the thickness of the spring, and the number of coils in the spring. The spring constant can also change if the spring is stretched or compressed beyond its elastic limit, causing permanent deformation.

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