Question about the conservation of mechanical energy

In summary, the mass sits at rest on the spring, so it is at the equilibrium position and thus mg = kd. When the mass at rest at the top. Its KE and PE is 0. When the mass at distance D, the question said the mass is stopped, its KE is 0, PE is mg(-D), elastic energy is 1/2k(-D) ^2. By conservation of energy: 0 = -mgD + 1/2kD^2. D = 2mg/k = 2d.
  • #1
TkoT
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Homework Statement
when a mass sits at rest on a spring, the spring is compressed by a distance d from its undeformed length(Fig.a). Suppose instead that the mass is released from rest when it barely touches the undeformed spring(fig.b). Find the distance D that the spring is compressed before it is able to stop the mass. Does D = d? if not why not?
Relevant Equations
F=-kx
conservation of energy
For the first part, the mass sits at rest on the spring, so it is at the equilibrium position and thus mg = kd
So, d = mg/k

For the second part, I assume the uncompressed spring position is 0. When the mass at rest at the top. Its KE and PE is 0. When the mass at distance D, the question said the mass is stopped, its KE is 0, PE is mg(-D), elastic energy is 1/2k(-D) ^2

By conservation of energy:

0 = -mgD + 1/2kD^2

D = 2mg/k = 2d

Question: I can get the answer correct but do not know why they are different. For the second part, I am not clear about what is the whole process happening. When the mass is released, the spring is compressed. without external force, I guess the mass will undergo SHM or just reach the equilibrium position like the first part. I have read the answer, the answer said there is external force like hand to cause this difference. I just can't figure out how the external force acting on the second part. Is it because the question said to stop the mass that mean external force is needed to hold the mass at the compressed position without rebounding? If no external force apply, is my guess correct?
 

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  • #2
The equilibrium position is where there is no net force on the mass. The mass will therefore undergo SHM around that point and since the amplitude on either side of that point is the same, the second turning point will be when the spring is compressed two times as much as in the equilibrium.

Difficult to state anything about the provided answer without having a verbatim copy.
 
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  • #3
Orodruin said:
The equilibrium position is where there is no net force on the mass. The mass will therefore undergo SHM around that point and since the amplitude on either side of that point is the same, the second turning point will be when the spring is compressed two times as much as in the equilibrium.

Difficult to state anything about the provided answer without having a verbatim copy.
thank you for your reply.
for the second turning point, is it because of the external force so it is two times larger? if not external force, is the second turning point equal to the first one?
i attached the answer
 

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  • #4
The point is that the "mass" is moving when it passes through the equilibrium position (the position where the net force is zero even though the velocity is nonzero).
 
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  • #5
TkoT said:
Question: I can get the answer correct but do not know why they are different. For the second part, I am not clear about what is the whole process happening. When the mass is released, the spring is compressed. without external force, I guess the mass will undergo SHM or just reach the equilibrium position like the first part. I have read the answer, the answer said there is external force like hand to cause this difference. I just can't figure out how the external force acting on the second part. Is it because the question said to stop the mass that mean external force is needed to hold the mass at the compressed position without rebounding? If no external force apply, is my guess correct?
Regarding Figure a:
Your guess is correct, once released to fall over the unloaded spring, the mass will have no initial resistance from the it (Weight of block > Spring force at zero deformation) and will initiate a free fall (See position 1 in diagram).

As the upwards resistance of the spring increases, that freedom of the block to fall decreases until any further downwards movement is being fully restricted by the maximum force reached by the spring (See position 2 in diagram).
At that point, the weight of the block is insufficient to counteract the max spring force, and it is accelerated upwards (Weight of block < Spring force at max deformation).
A simple harmonic oscillation respect to a midway equilibrium point is stablished (See position 3 in diagram).

That oscillation could last forever, but internal friction of the spring metal and air drag will gradually consume mechanical energy until reaching a minimum value (Ek tends to zero).
That remaining value is the potential gravity energy (Ep) of the block at that equilibrium point of zero velocity and zero acceleration (Weight of block = Spring force at half of previous max deformation).

Regarding Figure b:
The block is now not free to fall over the unloaded spring.
The spring is slowly and precisely pre-loaded until reaching the equilibrium point of zero velocity and zero acceleration by an external force (hand in this example) (See position 3 in diagram).

The block is carefully deposited over the pre-loaded spring and remains at that point of equilibrium of forces (downwards weight and upwards pre-loaded spring force), which results in zero acceleration (a=Fnet / m).

Since v=0, at this balance point, there is no extra kinetic energy to get degraded by friction and drag, all we see is potential elastic energy (Epe) of the spring precisely counteracting the potential gravity energy (Ep) of the block.
Block on spring.jpg
 

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  • #6
Love the title "conversation of energy", which is basically what is happening in this thread :)
 
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  • #7
Lnewqban said:
Since v=0, at this balance point, there is no extra kinetic energy to get degraded by friction and drag, all we see is potential elastic energy (Epe) of the spring precisely counteracting the potential gravity energy (Ep) of the block.
I think what you want to say here is that the upward elastic force counteracts the downward force of gravity. Kinetic and potential energy don't usually counteract each other considering that their zeroes are chosen arbitrarily.
 
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  • #8
kuruman said:
I think what you want to say here is that the upward elastic force counteracts the downward force of gravity. Kinetic and potential energy don't usually counteract each other considering that their zeroes are chosen arbitrarily.
Much better worded your way.
Correction welcome.
Thank you.
 
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1. What is the law of conservation of mechanical energy?

The law of conservation of mechanical energy states that the total amount of mechanical energy in a closed system remains constant over time, as long as there is no external work or non-conservative forces acting on the system.

2. What is mechanical energy?

Mechanical energy is the sum of kinetic energy and potential energy in a system. Kinetic energy is the energy of motion, while potential energy is the energy stored in an object due to its position or configuration.

3. How is mechanical energy conserved?

Mechanical energy is conserved because it cannot be created or destroyed, only transferred or transformed. In a closed system, the total amount of mechanical energy will remain constant, even as it changes form from kinetic to potential energy and vice versa.

4. What is an example of the conservation of mechanical energy?

A common example of the conservation of mechanical energy is a pendulum. As the pendulum swings back and forth, its kinetic energy is constantly changing into potential energy and back again, but the total amount of mechanical energy remains constant.

5. Are there any exceptions to the conservation of mechanical energy?

Yes, there are a few exceptions to the conservation of mechanical energy. In real-world systems, some energy is always lost to friction and other non-conservative forces, resulting in a decrease in the total mechanical energy over time. Additionally, in quantum mechanics, the principle of uncertainty allows for temporary violations of energy conservation on a very small scale.

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