Conservation of Energy vs Force Analysis for Springs

[fee6]

Homework Statement

A stone of mass $$m$$ is at rest on a vertical spring which is compressed a distance $$x$$. Find its spring constant $$k$$.

All variables are given in the problem.

Homework Equations

I solved this problem realizing that because the mass is at equilibrium, the sum of the vertical forces on it is 0. Thus, the force of gravity downwards is equal to the force of the spring upwards, and so $$mg = kx$$.

The Attempt at a Solution

Therefore, $$k = \frac{mg}{x}$$, from the previous equation. The problem, now, is that after solving it like this, a friend asked me why we could not use the conservation of energy:

$$E_i = E_f$$
$$mgx = \frac{1}{2}kx^2$$
$$k = \frac{2mg}{x}$$

which differs from the previous solution. I am inclined to believe that there is something wrong with the energy method, but I cannot put my finger on it. I would greatly appreciate any help with this. Thanks!

 

[PhanthomJay]

Homework Statement

A stone of mass $$m$$ is at rest on a vertical spring which is compressed a distance $$x$$. Find its spring constant $$k$$.

All variables are given in the problem.

Homework Equations

I solved this problem realizing that because the mass is at equilibrium, the sum of the vertical forces on it is 0. Thus, the force of gravity downwards is equal to the force of the spring upwards, and so $$mg = kx$$.

The Attempt at a Solution

Therefore, $$k = \frac{mg}{x}$$, from the previous equation. The problem, now, is that after solving it like this, a friend asked me why we could not use the conservation of energy:

$$E_i = E_f$$
$$mgx = \frac{1}{2}kx^2$$
$$k = \frac{2mg}{x}$$

which differs from the previous solution. I am inclined to believe that there is something wrong with the energy method, but I cannot put my finger on it. I would greatly appreciate any help with this. Thanks!

Your first method is correct. The energy method assumes that the block is released from rest at the unstretched position of the spring, in which case the mass comes to a stop at a distance twice the 'x' distance (it will oscillate back and forth and ultimately damp out and settle in its equlibrium position). The problem did not assume this; the mass is gently lowered by your hand until it reaches it's equilibrium position.

 

[fee6]

Your first method is correct. The energy method assumes that the block is released from rest at the unstretched position of the spring, in which case the mass comes to a stop at a distance twice the 'x' distance (it will oscillate back and forth and ultimately damp out and settle in its equlibrium position). The problem did not assume this; the mass is gently lowered by your hand until it reaches it's equilibrium position.

Ah, I see. If the mass were just dropped, which would cause it to come to a stop at twice the 'x' distance, would the equilibrium point then be at 'x' below the original position, and the amplitude of the resulting oscillations 'x' as well?

 

[PhanthomJay]

Ah, I see. If the mass were just dropped, which would cause it to come to a stop at twice the 'x' distance, would the equilibrium point then be at 'x' below the original position, and the amplitude of the resulting oscillations 'x' as well?

Yes, the equilbrium point (where the sum of forces = 0 ) is the same (at x below the unstretched length) in either case. The spring is not in equilibrium at '2x' if it were dropped, because even though it has temporarily come to a stop, it is still decelerating, net force is not 0 at that point. Good observation.