Calculating the spring force constant K

  • #1
lingo
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2
Homework Statement
A Spring launched cannon consists of a massless platform atop a vertical spring of elastic constant ##k## (top left). The spring must be compressed a distance ##h## (from its neutral position) to engage a latch mechanism that holds the platform in place until launch.

A block of mass m is gently placed on the platform, but the spring only compresses a distance ##h/3## before coming to equilibrium, and thus cannot lock into place (top right). To fully compress the spring and lock the mechanism, you must therefore push the block down with an external applied force of magnitude ##P##. Assume that you start pushing the moment you place the block (bottom left) and continue with a constant force until the latch engages (bottom right).

What is the smallest value for P that will allow the lock to engage? Express your answer as a multiple of the block's true weight, ##mg##.

Hint: You may assume the block ends at rest, at the moment the latch engages.
Relevant Equations
##\Delta E = W_{external}##
##F_s = kx##
So, my question is pertaining more to a specific part of this problem than actually calculating ##P## which I get to be ##P = \frac{kh}{2} - mg##. But I need ##P## in terms of a multiple of ##mg## so I need to find ##k##.

The solution attached uses the fact that when the object comes to a rest on the spring and is not moving, the downward force acting on the block, ##mg##, is equal to the spring force at that compression distance ##F_s = \frac{kh}{3}##. Then ##k = \frac{3mg}{h}##.

However, my solution involves the conservation of energy. Isn't the amount of work done by only the gravitational force, ##mg##, without the involvement of the external force ##P##, equal to the amount of potential energy stored in the spring. So then ##mg(\frac{h}{3}) = \frac{1}{2}k(\frac{h}{3})^2##.

Then solving for ##k## gives ##k = \frac{6mg}{h}##.

I dont know which ##k## is correct; is my approach incorrect?
 

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  • #2
Your energy conservation argument is wrong. For that to be the case the weight would be released from rest at ##y=0##. It would be oscillating in that scenario. This is not what happens though, a hand gently places the mass onto the platform. It’s sitting in equilibrium initially before the application of ##P##.
 
  • #3
lingo said:
A block of mass m is gently placed on the platform, but the spring only compresses a distance ##h/3## before coming to equilibrium, … you [then] push the block down with an external applied force of magnitude .
This is unclear. You could read it that the block is gently placed atop the uncompressed spring, then released, and after descending h/3 it passes through the equilibrium position, continuing down under its momentum while you add a force P.
(Even then, it does not match your analysis because you interpreted "at equilibrium" as meaning "instantaneously stationary".)
However, it almost certainly intends that the block is placed on the spring then gently lowered to the equilibrium position.
 
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  • #4
The hand exerts a non-conservative force on the block. Therefore, if you want to use mechanical energy conservation, you must leave the hand out of the picture entirely. In that case, the only way to lock the spring at distance ##h## below the neutral position, is by using the conservative force of gravity. This would require changing the statement of the problem to

A spring launched cannon consists of a massless platform atop a vertical spring of elastic constant ##k##. The spring must be compressed a distance ##h## (from its neutral position) to engage a latch mechanism that holds the platform in place until launch. When a block of mass ##m## is gently placed on a platform of negligible mass, the spring only compresses a distance ##h/3## before coming to equilibrium.

Find the minimum height ##y## above the platform at which the block must be released from rest so that the spring is locked at distance ##h## below the neutral position."

Minimum ##y## means that the block reaches distance ##h## with zero kinetic energy. Then using mechanical energy conservation from the point of release to point ##h## below the neutral position, $$\begin{align} &\Delta K+ \Delta U_{gravity}+\Delta U_{elastic}=0 \nonumber \\
& 0-mg(y+h)+\frac{1}{2}kh^2=0. \nonumber \end{align}$$Using ##k=\dfrac{3mg}{h}##, the answer is ##y=\frac{1}{2}h.##
 
  • #5
erobz said:
Your energy conservation argument is wrong. For that to be the case the weight would be released from rest at ##y=0##. It would be oscillating in that scenario. This is not what happens though, a hand gently places the mass onto the platform. It’s sitting in equilibrium initially before the application of ##P##.
I'm not sure I'm understanding you. Isn't the block indeed released from rest? I interpret the problem being that the block is placed on the spring, and the weight of the block compresses the spring a distance of ##\frac{h}{3}## until the spring force now, ##k\frac{h}{3}##, equals the gravitational force, ##mg##, and the block stays at rest there but are you saying the block won't stay at rest there and instead will oscillate back and forth? I'm not sure I'm interpreting the problem correctly.

This is all happening without the involvement of ##P##. In the attached image, there's two diagrams, the first image doesn't show any ##P## and the block compresses the spring a distance of ##\frac{h}{3}## while the next one shows ##P## applied right from the start before the block even travels a distance of ##\frac{h}{3}##.
 
  • #6
lingo said:
I'm not sure I'm understanding you. Isn't the block indeed released from rest? I interpret the problem being that the block is placed on the spring, and the weight of the block compresses the spring a distance of ##\frac{h}{3}## until the spring force now, ##k\frac{h}{3}##, equals the gravitational force, ##mg##, and the block stays at rest there but are you saying the block won't stay at rest there and instead will oscillate back and forth? I'm not sure I'm interpreting the problem correctly.

This is all happening without the involvement of ##P##. In the attached image, there's two diagrams, the first image doesn't show any ##P## and the block compresses the spring a distance of ##\frac{h}{3}## while the next one shows ##P## applied right from the start before the block even travels a distance of ##\frac{h}{3}##.
I believe the hand (i.e. you) are setting the mass on top of the plate, but you are not immediately releasing it. You are doing work on it so that it comes to static equilibrium position ( without kinetic energy) compressing the spring ##h/3##. You ( the hand) have removed all the kinetic energy it would have otherwise had moving from ##y=0## to ##y = -h/3## under the influence of gravity and the spring alone. If you let it go, it stays put. Then you are meant to apply ##P## to compress the spring further to the latch. I think the diagram is poorly drawn.

If you (the hand) just let the mass go at ##y=0## in the first diagram, it would accelerate, blow past the static equilibrium position (##y= -h/3##) and come to a stop below it (instantaneously) and return to its initial position (instantaneously) where you let it go. Thats what I meant by "it would oscillate". It would just keep doing that in the absence of non-conservative work.
 
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  • #7
Now looking again at the original problem statement, I'm not sure what is going on. "Assume you start pushing the moment you place the block", has thrown me for a loop. We are going to accelerate the block, under constant force ##P## such that it comes to rest at ##-h## I guess? Seems like that would result in solving the ODE ## \uparrow^+##

$$ -P - kx - mg = m\ddot x $$

? Might need a pro to set the record straight. At first I thought it was an unclear worded more simple problem...now I'm not sure.

EDIT: Never mind about the ODE, we just still utilize Work-Energy. Set PE = 0 at ## y = 0 ##. This time account for ##P## doing work. You get ## k## from sum of the forces = 0. We should be able to find ##P##.
 
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  • #8
erobz said:
"Assume you start pushing the moment to place the block", has thrown me for a loop.
I missed that too.
The wording is misleading. What they mean is "If you were simply to let the mass go, or lower it gently, it would eventually come to rest after descending h/3 (and without ever reaching the latch). So, instead of doing that, you push down with force P right from the start." The first part gives you k, the second part gives the energy equation.
 
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  • #9
erobz said:
$$ -P + kx - mg = m\ddot x $$
The ##kx## term on the LHS must have a negative sign in front of it because the force is restoring. In other words, if you set ##P=0## and "turn off" gravity by placing the spring-mass system on a horizontal frictionless plane, you should recover the simple harmonic oscillator ODE.
 
  • #10
haruspex said:
The wording is misleading. What they mean is "If you were simply to let the mass go, or lower it gently, it would eventually come to rest after descending h/3 (and without ever reaching the latch). So, instead of doing that, you push down with force P right from the start." The first part gives you k, the second part gives the energy equation.
I agree.
 
  • #11
kuruman said:
The ##kx## term on the LHS must have a negative sign in front of it because the force is restoring. In other words, if you set ##P=0## and "turn off" gravity by placing the spring-mass system on a horizontal frictionless plane, you should recover the simple harmonic oscillator ODE.
Oh yes, I didn't notice by my convention my displacement ##x## is negative. Thank you!
 
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  • #12
haruspex said:
I missed that too.
The wording is misleading. What they mean is "If you were simply to let the mass go, or lower it gently, it would eventually come to rest after descending h/3 (and without ever reaching the latch). So, instead of doing that, you push down with force P right from the start." The first part gives you k, the second part gives the energy equation.
Well, this was how I interpreted the first part. The mass would eventually come to a rest and the gravitational force, ##mg## would equal the current spring force ##k\frac{h}{3}## and you solve for k as was done in the attached solution. But this led me to believe that if you if let the mass go, isn't the work done by gravity, ##mg\frac{h}{3}## equal to the amount of potential energy stored in the spring when it compresses a distance of ##\frac{h}{3}## ? But this would of course give me a wrong ##k##, if I set up the equation ##mg\frac{h}{3} = \frac{1}{2}k(\frac{h}{3})^2##. Is it even possible to find k using work and energy in the first part? Or is it we just don't know the total work being done by external forces?
 
  • #13
lingo said:
this led me to believe that if you if let the mass go, isn't the work done by gravity, ##mg\frac{h}{3}## equal to the amount of potential energy stored in the spring when it compresses a distance of ##\frac{h}{3}## ?
No, you are ignoring KE. Suppose you let it go from the relaxed position. When it reaches the equilibrium position the spring force has only just come equal to mg, so until that point it was still accelerating. It is now at maximum speed
##mg\frac{h}{3}=\frac 12k(h/3)^2+\frac 12 mv_{max}^2##
and will continue down past the equilibrium point until
##mg\frac{2h}{3}=\frac 12k(2h/3)^2##.
 
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  • #14
haruspex said:
No, you are ignoring KE. Suppose you let it go from the relaxed position. When it reaches the equilibrium position the spring force has only just come equal to mg, so until that point it was still accelerating. It is now at maximum speed
##mg\frac{h}{3}=\frac 12k(h/3)^2+\frac 12 mv_{max}^2##
and will continue down past the equilibrium point until
##mg\frac{2h}{3}=\frac 12k(2h/3)^2##.
I see now. Thank you. I guess when I read the problem, I was thinking the mass would come to rest at ##\frac{h}{3}## but I was wrong, and I suppose the nature of the problem wouldn't allow that to happen if the spring force eventually equals the gravitational force after the mass travels some certain distance. Because if it somehow came to rest, well I guess at some point between ##0## and ##\frac{h}{3}##, ##F_s## would have to be greater than ##mg## and I also think that would make the mass go back up and do something else... oscillate?

Also, is that what the problem setter means by equilibrium, that there is no net external force acting on the mass? Thought it meant the mass is not moving.
 
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  • #15
lingo said:
is that what the problem setter means by equilibrium, that there is no net external force acting on the mass?
Yes.
"In classical mechanics, a particle is in mechanical equilibrium if the net force on that particle is zero.  By extension, a physical system made up of many parts is in mechanical equilibrium if the net force on each of its individual parts is zero."
https://en.wikipedia.org/wiki/Mechanical_equilibrium

lingo said:
Thought it meant the mass is not moving.
To be honest, I also thought it meant that, in addition to the above, the system is static. So I would refer to the equilibrium position if I meant where it would rest if it were not moving. Having read the Wikipedia definition, it seems the problem wording is fine.
 
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