Question about the conservation of mechanical energy

[TkoT]

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?

 

[Orodruin]

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.

 

[TkoT]

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

 

[Mister T]

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).

 

[Lnewqban]

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.

 

[malawi_glenn]

Love the title "conversation of energy", which is basically what is happening in this thread :)

 

[kuruman]

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.

 

[Lnewqban]

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.