Help with springs and potential energy problem

In summary, a massless platform supported by 4 equal springs with a mass m placed on the center causing an equal compression of x in each spring will result in a spring constant k of (mg)/(4x) due to the fact that mechanical energy is not conserved and an additional term must be accounted for in the equation.
  • #1
GeoMike
67
0
Ok, this seems simple, and it probably is, but I'm just not seeing why I'm getting two non-equivalent solutions...

Homework Statement


"A massless platform is supported by 4 equal springs (same length, same spring constant, etc.)
A mass m is placed on the center of the platform causing an equal compression of x in each spring. What is the spring constant k of a single spring in terms of the mass m, the acceleration of gravity g, and the compression of the spring x?

Homework Equations


Fspring = -kx
Fgrav = mg
Uspring = 1/2(kx^2)
Ugrav = mgx

The Attempt at a Solution


What's throwing me off is that there seem to be two valid ways to solve this, each resulting in a different value of k...

If I use the fact that the force acting on each spring is 1/4 the force of gravity acting on the mass I get:

kx=mg/4
So... k = (mg)/(4x)

But, then isn't the elastic potential energy stored in each spring 1/4 the gravitational potential energy in the mass when it is first placed on the platform (before any compression)? If so, I get:

1/2(kx^2) = mgx/4
So... k = (mg)/(2x)

The back of the book gives (mg)/(4x). Which I understand, the problem is that the second answer seems valid to me, but obviously the second one is wrong. I just can't see where is my error in reasoning. Help?

Thank you!
-GM-
 
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  • #2
GeoMike said:
Ok, this seems simple, and it probably is, but I'm just not seeing why I'm getting two non-equivalent solutions...

Homework Statement


"A massless platform is supported by 4 equal springs (same length, same spring constant, etc.)
A mass m is placed on the center of the platform causing an equal compression of x in each spring. What is the spring constant k of a single spring in terms of the mass m, the acceleration of gravity g, and the compression of the spring x?

Homework Equations


Fspring = -kx
Fgrav = mg
Uspring = 1/2(kx^2)
Ugrav = mgx

The Attempt at a Solution


What's throwing me off is that there seem to be two valid ways to solve this, each resulting in a different value of k...

If I use the fact that the force acting on each spring is 1/4 the force of gravity acting on the mass I get:

kx=mg/4
So... k = (mg)/(4x)

But, then isn't the elastic potential energy stored in each spring 1/4 the gravitational potential energy in the mass when it is first placed on the platform (before any compression)? If so, I get:

1/2(kx^2) = mgx/4
So... k = (mg)/(2x)

The back of the book gives (mg)/(4x). Which I understand, the problem is that the second answer seems valid to me, but obviously the second one is wrong. I just can't see where is my error in reasoning. Help?

Thank you!
-GM-
That is a very sharp observation on your part. The error in your reasoning is that mechanical energy is not conserved, since you must apply a non conservative force to slowly lower the beam to its new rest postion; otherwise it would osscilate back and forth about its final rest position, extending the spring twice as much in so doing.
 
  • #3
PhanthomJay said:
That is a very sharp observation on your part. The error in your reasoning is that mechanical energy is not conserved, since you must apply a non conservative force to slowly lower the beam to its new rest postion; otherwise it would osscilate back and forth about its final rest position, extending the spring twice as much in so doing.

So basically I forgot to account for the fact that if the mechanical energy is conserved (i.e. the mass is simply released, not slowly lowered) then an oscillation about x occurs requiring an additional term to account for the fact that there is a non-zero velocity as the mass passes through x (it has kinetic energy as well).

Right?
-GM-
 
  • #4
Right!
 

1. What is potential energy?

Potential energy is the energy that an object possesses due to its position or state. In the case of a spring, potential energy is stored when the spring is stretched or compressed.

2. How is potential energy related to springs?

Potential energy is related to springs through Hooke's Law, which states that the potential energy stored in a spring is directly proportional to the amount that the spring is stretched or compressed. This means that the more a spring is stretched or compressed, the more potential energy it has.

3. How do I calculate the potential energy of a spring?

The potential energy of a spring can be calculated using the equation PE = 1/2kx^2, where PE is potential energy, k is the spring constant, and x is the displacement of the spring from its equilibrium position.

4. What is the relationship between potential energy and kinetic energy in a spring?

In a spring, potential energy is converted into kinetic energy as the spring oscillates back and forth. At the maximum displacement, all of the potential energy is converted into kinetic energy, and at the equilibrium position, all of the kinetic energy is converted back into potential energy.

5. How does the mass of an object affect the potential energy of a spring?

The mass of an object does not directly affect the potential energy of a spring. However, a heavier object will require more force to stretch or compress the spring, resulting in a larger displacement and therefore a greater amount of potential energy stored in the spring.

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