The 2.5-kg wieght is released from rest -- Determine the spring constant k

In summary, the conversation is about determining the value of k for which a 2.5-kg weight, released from rest at position A with two undeformed springs of stiffness k, would reach position B. The conversation includes a discussion on calculating potential energy and the relevance of the zero point for potential energy. It is suggested that the problem be posted in the "Advanced Physics Homework" forum. A solution is provided but not fully calculated.
  • #1
Alexanddros81
177
4

Homework Statement


14.27 The 2.5-kg weight is released from rest in position A, where the two springs
of stiffness k each are undeformed. Determine the largest k for which the weight
would reach position B

Fig P14_27.jpg


Homework Equations

The Attempt at a Solution



Pytels_Dynamics115.jpg


Hi. Can you check if I am going at the right direction?
Also for ##Wy_B## and ##Wy_A## am I replacing with 2.5(0.45) and 2.5(0.3) respectively?
 

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  • #2
So far so good except for the gravitational potential energy.
 
  • #3
Do I replace ##Wy_A## and ##Wy_B## with 2.5(0.3) and 2.5(0.45) respectively?
 
  • #4
Alexanddros81 said:
Do I replace ##Wy_A## and ##Wy_B## with 2.5(0.3) and 2.5(0.45) respectively?
Almost but not quite. Something is missing. The problem talks about a 2.5-kg weight. Is weight measured in kilograms?
 
  • #5
A yes but ofcourse W = mg where mass is m=2.5kg.
But I don't understant why the Potential Energy of W is mg(0.3) at position A and mg(0.45) at position B.
I mean I don't understant it in relation to the Figure.
Can you explain to me a bit more?
At Which point do I say that the potential energy of W is zero? (in the figure)
 
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  • #6
Alexanddros81 said:
At Which point do I say that the potential energy of W is zero? (in the figure)
Any point you like. The zero of potential energy is irrelevant. Consider this, when a mass moves from point A to B, mechanical energy conservation says
TA + UA = TB + UB
or
TB - TA + UB - UA = 0
or
ΔT + ΔU = 0.
It's the difference that matters, not where you put the zero of energy because it drops out.
 
  • #7
By the way do I post problem questions from
'Orbital Mechanics for Engineering Students' by Curtis
in this forum (Introductory Physics Homework)?
 
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  • #8
Alexanddros81 said:
By the way do I post problem questions form Orbital Mechanics for Engineers by Curtis
in this forum (Introductory Physics Homework)?
I am not familiar with the textbook. I looked at its table of contents and I think "Advanced Physics Homework" would be more appropriate. Try that and if I am wrong, a mentor will move it to its rightful place, perhaps in "Engineering."
 
  • #9
Ok so here is my solution:

Pytels_Dynamics118.jpg
 

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  • #10
I didn't run the numbers, but the method is correct.
 
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Likes Alexanddros81

Related to The 2.5-kg wieght is released from rest -- Determine the spring constant k

1. What is the equation used to determine the spring constant k?

The equation used to determine the spring constant k is k = mg/x, where m is the mass of the object, g is the acceleration due to gravity, and x is the displacement of the spring.

2. How does the mass of the object affect the spring constant k?

The mass of the object has a direct effect on the spring constant k. As the mass increases, the spring constant also increases, meaning the spring becomes stiffer and requires more force to stretch it.

3. What is the unit of measurement for the spring constant k?

The unit of measurement for the spring constant k is N/m (newtons per meter). This represents the amount of force required to stretch the spring by 1 meter.

4. Can the spring constant k change?

Yes, the spring constant k can change depending on factors such as the material of the spring, the temperature, and the amount of strain the spring has experienced. However, for a specific spring, the spring constant remains constant as long as the spring follows Hooke's law.

5. How does the acceleration due to gravity affect the spring constant k?

The acceleration due to gravity, g, has a direct effect on the spring constant k. As g increases, the spring constant also increases, meaning the spring becomes stiffer and requires more force to stretch it.

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