Why Do Force and Energy Analyses Yield Different Spring Extensions?

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Homework Help Overview

The discussion revolves around a system involving a block of mass M attached to a spring on an inclined plane, specifically focusing on the differing results obtained from force and energy analyses when calculating the maximum extension of the spring.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to use both energy conservation and force balance equations to find the spring extension, noting discrepancies in the results. Some participants question the application of forces in the energy conservation context and suggest clarifying the situation with a diagram. Others raise concerns about the treatment of potential and kinetic energy in the analysis.

Discussion Status

The discussion is exploring various interpretations of the problem, with participants offering different perspectives on the energy conservation approach and the role of forces in the system. There is no explicit consensus, but several lines of reasoning are being examined.

Contextual Notes

Participants are addressing potential misunderstandings regarding the definitions of work and energy in the context of the problem, as well as the assumptions made about the system's dynamics, such as the acceleration of the mass and the relationship between height and spring extension.

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I have a question about thse following system: block of mass M is attached to a spring is on an inclined plane.
Why do the force and energy analysis equations give different answers? The question is to find the maximum extension of the spring. Here is my thinking so far.


I know if you use energy convervation, you get 1/2*k*x^2 = M*g*sin(theta), so you get x = 2*M*g*sin(theta)/k.
However, why can I not just set the force along the incline equal to the spring force and solve for x that way? If I do that, I get a different answer:
k*x = M*g*sin(theta)
x = M*g*sin(theta)/k and this is different by a factor of 2.
 
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What exactly is the situation here? Could you draw a diagram?
What's with the mgsin(theta)? It's a force, not work. You cannot use it in the energy conservation equation.
 
The problem is the energy convervation. The potential energy should be
M*g*h were h is the height difference of the two states (before the mass was released and after).
What about kinetic energy as the weight get to the end? (it has mass and it is moving)
Also, the way that you did the first answer is missing a square route.
If you can find the connection between the h and the x then the ^2 could cancel out!
 
In the answer above I assumed that you wanted the equilibrium. If you want the maximum extension the you can't say that the force of gravity and the spring along the plane are equal because as the mass is accelerating as it gets to the bottom (harmonic motion)
 
A spring with constant k, and a displacement (x) from static deflection position (dst), also called equilibrium position.

Force: k*dst = W_block = mg
mg - k(x+dst) = ma
ma + kx = 0

ma here gives the force on a spring that expands it by x from equilibrium position, dst. F = -kx, by convention direction of x is negative to the dst, so F = kx

When you compress the spring you are doing mechanical work on it W = (F*dr) and storing energy (KE, PE, or both) in the spring. When the force is removed the spring does work on the block to move it a distance d against external forces. When it is not compressed it has no stored energy and a force is applied to expand the spring from its equilibrium position.
 

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