Convert Elastic Potential to Gravitational Potential

[Dr Gerrard]

---

Members don't see this ad.

All right, so there is a spring on the ground.


Two boxes are on top of the spring, compressing it. One box is twice as heavy as the other box.

When the spring is released, how high will the boxes go?

I said box 1 will go half the distance because it is twice as heavy as box 2.

The correct answer is that both will go the same height.

I used conservation of energy.

The initial elastic potential energy is the same for both boxes --- 1/2 kx^2

However, when the boxes leave the spring, this energy is converted to gravitational potential energy.

Since box 2 has a higher mass, it will have travel to a smaller height, because u = mgh.

Both boxes had the same elastic potential energy so they must also have the same gravitational potential energy, because this is the conservation of energy.

Where am I wrong?

The only thing I can think of is that the heavier box had a greater elastic potential energy, but I cannot really justify this.

 

[Dr Gerrard]

.

 

[Dr Gerrard]

With the falling ball examples, gravitational potential energy is converted to kinetic energy, both of which take mass into consideration so mass can be cancelled in both of those.

Additionally, changing the mass only changes to force due to gravity, but the force changes as to keep acceleration constant.

But still, I am not really understanding the springs. When two boxes are on a spring, does each one have a certain amount of elastic potential or are they both the same?

Or maybe, since they are both on the same spring, they travel together?

But still, if you have two identical springs, both compressed the exact same amount. One has a 2kg box while the other has a 1kg box, wouldn't the 1kg box travel twice the distance in the air in order to conserve energy?

 

[Bernoull]

With the falling ball examples, gravitational potential energy is converted to kinetic energy, both of which take mass into consideration so mass can be cancelled in both of those.

Additionally, changing the mass only changes to force due to gravity, but the force changes as to keep acceleration constant.

But still, I am not really understanding the springs. When two boxes are on a spring, does each one have a certain amount of elastic potential or are they both the same?

Or maybe, since they are both on the same spring, they travel together?

But still, if you have two identical springs, both compressed the exact same amount. One has a 2kg box while the other has a 1kg box, wouldn't the 1kg box travel twice the distance in the air in order to conserve energy?

The only way I can justify the same height answer is from a purely kinematics viewpoint. Both objects have the same velocity at take-off (when spring's at equilibrium position), in the absence of frictional force/air resistance, the only thing decelerating the objects' velocity thus flight time and max height is gravity which is independent of mass.

I've tried conservation of mechanical energy approach:
KEi + GPEi + EPEi = KEf + GPEf + EPEf

but it yields the same answer u got..🙁🙁 Something's off, both approaches should yield the same result except if we're overlooking something or not making certain assumptions. I'm modelling the problem as a TWO-spring-mass-system rather than one..

 

[Dr Gerrard]

Similarly, the period of a pendulum is also not dependent on mass.

Even though T = 2(pi)(m/k)^0.5

This leads me to believe that something in K incorporates mass, but that does not make sense, because K is a property of the spring and does not change depending on the mass of the object m.

Thanks for your help though Bernoull, I also understand it slightly from the kinematics perspective, although I still would say that they do not both have the same initial velocity, because the heavier one will have a lower initial velocity because kinetic energy also has an "m" in it.

 

[fishbait63]

I think the problem is that you are considering that they would have the same force of compression on the spring.

If m1 (mass1) is on top of spring it applies mg down on the spring and the spring has force -k* dx

if m2 is half the weight of mass1 it only applies 1/2mg and the spring only has force -.5k *dx

so mass1 will compress it twice as much as m2

m1 has potential energy .5k*x^2
m2 has pe .5k (.5x)^2

I hope this helps somehow

 

[Dr Gerrard]

Yeah, but the thing is, both boxes are on the same spring, compressing it at the same time.

How does that affect the situation?

If a 10 kg block and a 5 kg block were both placed on the SAME spring, and the spring were compressed, how would their height differ?

EK says it wouldn't. I don't understand why, when dealing with conservation of energy.

 

[Bernoull]

Similarly, the period of a pendulum is also not dependent on mass.

Even though T = 2(pi)(m/k)^0.5

This leads me to believe that something in K incorporates mass, but that does not make sense, because K is a property of the spring and does not change depending on the mass of the object m.

Thanks for your help though Bernoull, I also understand it slightly from the kinematics perspective, although I still would say that they do not both have the same initial velocity, because the heavier one will have a lower initial velocity because kinetic energy also has an "m" in it.

.
First let's clarify a few things:
1. By take-off velocity I meant maximum velocity of spring-mass system. This is the only velocity you can use for conservation of energy IF you equate Potential E and Kinetic E (BOTH MUST be maximised). Max velocity occurs at spring's equilibriumum position and it equals spring recoil velocity at that point. If you picture a board on top of the spring on which both masses sit it's easy to see that both masses move at the same velocity as the board.

2. Past the equilibrium position of spring, its recoil velocity decreases due to hooke's force and the masses continue at max velocity so they take-off - (analogous to sitting on da roof of ur friend's car as it goes 60mph and he slams the brakes, well u keep moving at 60mph (with no air resistance)). Take off velocity is the same for both masses and then they decelerate by 10m/s^2 and thus they have the same flight time and max height...

They same result can be arrived at from a conservation of energy perspective, the problem we have had is twofold: first we didn't appreciate the dependence of k on mass and second we didn't correctly define a system to analyse.

conservation of mechanical energy approach: (assuming no rotation)
KEi + GPEi + EPEi = KEf + GPEf + EPEf

We can set max EPE = Max KE = Max GPE
0.5kx^2 = 0.5mv^2 = mgh

Simplify to Max KE = Max GPE
0.5mv^2 = mgh

NOW I THINK WE NEED TO BE ABSOLUTELY CAREFUL IN DEFINING OUR SYSTEM. I'll define two systems considering each mass as a seperate system. Mass 1 = 2m and Mass 2 = m

Max KE (Mass 1) = Max GPE (Mass 1)
0.5*2mv^2 = 2mgh, 2m(s) cancel out and h=0.05v^2 where v = max velocity

Max KE (Mass 2) = Max GPE (Mass 2)
0.5*mv^2 = mgh, m(s) cancel out and h=0.05v^2 where v = max velocity

CLEARLY max heights are equal, but u gotta separate out the masses and analyse them independently, if you calculate Max KE by summing up the masses, then u gotta sum up the masses for Max GPE also. Consistency is crucial here.

Now looking at the elastic constant:

F=kx (deliberately ignoring the - sign)
N = kg*m/s^2 = k*m, thus dimensionally k =kg/s^2

This is key because superficically hooke's force or EPE seems to be independent of mass, but by experience we know a heavier mass will compress a spring more so this can't be. If you do the analysis between EPE and GPE, independently for both masses as b4:

max EPE = max GPE
0.5kx^2 = mgh
0.5*kg*m^2/s^2 = kg*m^2/s^2 since masses are equal on both sides, they cancel.

0.5v^2 = gh and h= 0.05v^2

All of this simply says that for springs, maximum height for an object that "takes flight" is independent of mass because KE, GPE, EPE all account for the object's mass and they cancel out!!!


I will rest my case... .

 

[Dr Gerrard]

Wow, thanks Bernoull for writing all of that out.

One more question though.

Since k in the elastic potential energy equation does not change with different masses, how does it account for different values of m in the gravitational potential energy equation? Is this just an inherent property of k and am I looking to get into much more information than needed?

 

[Bernoull]

Wow, thanks Bernoull for writing all of that out.

One more question though.

Since k in the elastic potential energy equation does not change with different masses, how does it account for different values of m in the gravitational potential energy equation? Is this just an inherent property of k and am I looking to get into much more information than needed?

I think u gotta analyse the masses separately/independently. Define ur system. For EPE what's ur system and for GPE what's ur system? If ur consistent in defining ur system, the masses cancel out from both EPE and GPE.. let me know if this is not clear..

 

[Dr Gerrard]

Well what would be the systems in the problem I posted in order for the masses to cancel out? It seems as though the mass associated in the spring would be the mass of both boxes while each box has a separate, smaller mass.

 

[Bernoull]

Yeah, but the thing is, both boxes are on the same spring, compressing it at the same time.

How does that affect the situation?

If a 10 kg block and a 5 kg block were both placed on the SAME spring, and the spring were compressed, how would their height differ?

EK says it wouldn't. I don't understand why, when dealing with conservation of energy.

Here's 2 examples of wat I mean by being consistent in defining ur system.

METHOD A:
If your system for EPE is BOTH masses, you're essentially modelling the masses as a point mass, in other words ur considering the center of the mass of the the two masses, and to simplify, u might as well pretend u had a SINGLE 15kg object. Calculating the EPE in this case, m=15kg.

Now to remain consistent, when you find GPE, u MUST use the same system (otherwise ur comparing oranges to apples) and GPE=mgh where m = 15kg. This method seems trivial bcos ur essentially not CONTRASTING the EPEs & GPEs of the two separate objects, and the answer is obvious.... masses cancel

METHOD B:
1. Here I'm defining TWO INDEPENDENT systems. One for each object.

10kg mass
EPE .. I find EPE due ONLY to the 10kg, that is my first system so I ignore everything else INCLUDING the 5kg mass.
For GPE of the 10kg system, well this is straightforward and obviously I'll only consider the 10kg for mass in GPE.. masses cancel out again...

2. Essentially repeat (1) but this time my system is the 5kg mass so both GPE and EPE, I'll use 5kg, thus they cancel also..


A more intuitive analogy is this: if I have $10 to split in some proportion b/t 2 kids, where each kids receives >$0, then each kid NECESSARILY has <$10 to spend..

In the same way, the 2 masses have a max EPE, which for argument purposes, let's hypothesize that it's 100J:

100J = .kg*m^2/s^2 and the mass here = 15kg, you CAN'T ascribe this 100J value to EACH 10kg and 5kg mass, doing so implies you somehow doubled the max EPE, BUT HOW, you can't create energy. Much like the $10 analogy, the 100J has to be shared by BOTH masses, so it's better to ONLY consider the energy each mass contributes to EPE, when calculating EPE, this way the masses cancel out


Hope this helps...
.