Finding x for a spring using energy equations

In summary, the conversation discusses finding the distance a spring stretches when a mass is hung on it, using both force equations and energy equations. While setting the spring force equal to the weight of the mass can solve for the distance, using energy equations may result in a different answer as it calculates the farthest the mass will fall, not the equilibrium position.
  • #1
cbasst
33
0

Homework Statement



This isn't a homework question, just something I've been thinking about. If given a situation in which mass m is hung on a spring with constant k, I know you can set the spring force equal to the weight of the mass to find how far it stretches. Is there a way to find how far it stretches using energy equations? It seems like it should be doable, but I can't figure out how.

Homework Equations



w = mg
F = -kx
ΔUg = mgh
Us = kx2/2

The Attempt at a Solution



As I said, setting mg = kx can solve for x. How do I do it with energy though?
Ug = Us
mgx = kx2/2

But now I get a different answer for x. What did I do wrong with the energy equations?
 
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  • #2
The energy approach will give you the farthest the mass on the spring will fall. This is not the equilibrium position. Think about it, if I let the mass drop, then when it passes the equilibrium position, it will not have zero velocity.
 
  • #3
Ah. This makes sense now! Thanks!
 

1. How do you find the spring constant (k) for a given spring using energy equations?

To find the spring constant, you can use the equation: k = (1/2) * (m * g) * (x/f)^2 where m is the mass of the object attached to the spring, g is the acceleration due to gravity, x is the displacement of the spring, and f is the frequency of oscillation. This equation is derived from the potential energy equation for a spring, PE = (1/2) * k * x^2.

2. What is the relationship between the displacement of a spring and its potential energy?

The displacement of a spring is directly proportional to its potential energy. This means that as the displacement increases, the potential energy also increases. This relationship is described by the equation: PE = (1/2) * k * x^2, where k is the spring constant and x is the displacement.

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

The mass of an object attached to a spring affects its energy by changing the frequency of oscillation. The greater the mass, the lower the frequency and therefore the less energy the spring has. This can be seen in the equation: E = (1/2) * m * w^2 * A^2, where E is the energy, m is the mass, w is the angular frequency, and A is the amplitude of oscillation.

4. Can you find the displacement of a spring if the energy and spring constant are known?

Yes, you can use the equation x = sqrt(2 * E / k) to find the displacement of a spring. This equation is derived from the potential energy equation for a spring, PE = (1/2) * k * x^2.

5. How do you calculate the force exerted by a spring using energy equations?

The force exerted by a spring can be calculated using the equation: F = -k * x, where k is the spring constant and x is the displacement of the spring. This equation is derived from Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement.

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