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### Course: Physics archive > Unit 5

Lesson 2: Springs and Hooke's law# Vertical springs and energy conservation

In this video, David explains two different strategies to deal with vertical springs and compares them with those used for horizontal springs. Created by David SantoPietro.

## Want to join the conversation?

- In his equation of summation of the forces acting during the equilibrium point. why isn't the force exerted by the spring -kx?(12 votes)
- great question . you can write in this way : -kx + mg = 0 or kx - mg = 0 . It's same . the negative sign is used to make the spring force exerted in opposite direction (
**gravitational force***mg*vs**spring force**-*kx*) I hope you got it. ;)(7 votes)

- What happens if you attach a mass to two vertical springs with different spring constants?(2 votes)
- it depends how you attach the springs... Parallel or series.(5 votes)

- Suppose instead of a spring it is a sling shot ( same idea) but instead of a vertical launch the launch is at some angle. If I wanted to work through conservation of energy issues to solve for the maximum height obtained for say the stone launched by the sling shot their would be a "horizontal" component of energy ( kinetic energy) to consider. That doesn't disappear.(3 votes)
- Nice thinking.

Think, therefore, where does the energy go?

Seems to me like you have kinetic energy horizontally (which remains constant

and increasing gravitational potential energy vertically

what you think?(2 votes)

- What happens if the string itself is being raised with a constant acceleration?.(3 votes)
- When we let go of the spring, is the force still there? If yes, what is it equal to?

If there was a block attached to the spring, would the spring exert a tension force to the block?(2 votes)- It is equal to the mass times the instantaneous acceleration of the block. In case there is no block attached, then the mass will be equal to that of the spring itself which shall accelerate (but different points will accelerate differently and shall have different tensions, a very complicated case). Much simpler to understand if we attach a block to a theoretically massless spring.(2 votes)

- So the kinetic energy of the mass is equal to the spring energy when we pull/push the mass and the spring pushes/pulls is back and when the mass is at equilibrium position?(1 vote)
- The kinetic energy of the spring is equal to its elastic potential energy, i.e. 1/2mv^2 = 1/2kx^2 when the spring is stretched some distance x from the equilibrium point and when its mass also has some velocity, v, with which it is moving. This occurs somewhere in between the equilibrium point and the extreme point (extreme point is when x=amplitude, A).

At the equilibrium, the spring is not stretched any distance away from the equilibrium, i.e. x=0 and thus the mass moves with maximum velocity (as the total energy = kinetic energy + elastic potential energy, and this is conserved). Therefore, at equilibrium, total energy = elastic potential energy.

At the extreme point, the spring is stretched to its max ability as x=A, and the velocity is zero as this is the point just before the velocity is about to change its direction. Therefore, at this point, the total energy = kinetic energy.

I hope that answered your question :)(4 votes)

- i have to find out if the mechanical energy is conserved in an oscillating spring mass apparatus (basically oscillating a mass hanging from a spring. What should be my initial and final mechanical energies? I'm thinking my initial mechanical energy will be when I pull the mass, and my final energy will be when the mass comes to rest after oscillation. Is that right? Can I find out if the energy is conserved that way?(2 votes)
- If it comes to rest after having been moving, was mechanical energy conserved?(2 votes)

- if i stretched a spring and let it go (assuming no energy is lost to heat ,sound...)would it go back and forth forever?(1 vote)
- Wouldn't gravity cause it to stay down after a long period of time and does't air resistance play a part that would eventually slow down the object enough to bring it to a stop. Even though air resistance barely constitutes to a substantial force it should still play a part in bringing it to a stop(3 votes)

- How do I solve a problem that pertains to a spring sitting on a horizontal plane when a 2kg mass is placed on top of it? I'm given that K= 100N/m and that the spring is compressed .15m, and I need to find the potential energy stored in the spring as well as the maximum height the object ascends.(1 vote)
- the spring is vertical right?

the mass is sitting on top the spring right? Is the mass causing it to compress? or is the spring compressed After the mass has been placed on it?

there is an equation for energy contained in a spring... you know it?

can you say why the object would ascend?(3 votes)

- So technically, if the block were to just be attached to the spring with someone holding the spring at its original length, and then they let go... The spring would then stretch to a maximum length, oscillate, and come to a new equilibrium position right?(2 votes)
- That sounds right. When the mass is attached to the spring, even though it is being held at the original length, it is still displaced from the equilibrium position, which in this case, is when the force of the spring is balanced out by the force of gravity on the mass (kx = mgh)(1 vote)

## Video transcript

- [Instructor] Let's say you've got a mass connected to a spring
and the mass is sitting on a frictionless surface. If the mass is sitting at a point where the spring is just at
the spring's natural length, the mass isn't going to go anywhere because when the spring
is at its natural length, it is content with its
place in the universe. It neither pushes nor pulls. It has no spring energy. This is like me most days in the summer. So we call this point where
the spring neither pushes nor pulls, the spring's natural length. For a mass on a horizontal spring, this is gonna be the equilibrium position. What we mean by equilibrium
position is the point where the net force on the mass is zero. So for a mass on a horizontal spring, the equilibrium position is at the point where the spring is at its natural length because the spring wouldn't be pushing to the right or the left. If you just put the mass
there at that point, it would just stay there forever at rest. That would be boring. So let's say we pull the mass to the right at distance d. If we do this, we give this
spring spring potential energy. If we release the mass from
rest while the spring has spring potential energy, the
spring's gonna pull the mass back to the left and
that mass is gonna move through the equilibrium
position with the sum speed. We can figure out what that speed is just by using conservation of
energy and it's not that hard. The potential energy, the
spring with start with would be 1/2 k, the
spring constant, times d, the amount of spring has
been stretched, squared. There would be no kinetic energy to start because we release the mass from rest. As this mass flies to the left, it would start gaining kinetic
energy, the spring energy would start turning into kinetic energy. When the mass gets to
the equilibrium position, the d would be zero. So at that point there
would be no spring energy and all of the spring
energy would have turned into kinetic energy and you
get this simple relationship that says all the spring energy equals all the kinetic energy at
the equilibrium position. So if we solve for v, we can
get that the v of this mass at the equilibrium position
would be the square root of k over m times d squared. You could call this d out because d squared and square root. but this is the idea. This is the speed you
would get of the mass passing through the equilibrium position. Let me ask you this. What if it was a vertical spring? And this mass is sitting here. We come find a vertical spring
with a mass hanging on it. We're like hey, I wanna pull this down, the distance d. If I pull it down to distance d, when this mass reaches the
equilibrium position again, will it also be going root
k over m times d squared? Or, would it be going
at some different speed because now it's hanging vertically? Well it turns the spring
constant's the same and you pull it down from the point where the mass is hanging. This exact same procedure
is gonna hold over here and you can find the speed
in the exact same way and that should be surprising. That should not be obvious. Because when the mass
is hanging over here, you don't just have a spring force. You got a gravitational force. You don't just have a spring
energy and kinetic energy. Think about it. This mass is moving up and down. You got changes and
gravitational potential energy. So why don't we have to taken into account the gravitational potential energy when we're doing conservation
of energy in this equation? Well that's what I wanna prove to you in the rest of this video. If all you wanted was
the result, if all you, if you're good, if you're like, "Man, all right, I can do
the same thing above cases, "don't even tell me anything else." You're good. But I suggest you watch the rest of it because knowing why you
can ignore the m g h in this calculation
gives you better insight into what we really mean by
this d and this h and this v as well as what we really mean
by the equilibrium position and that will conceptually aid you if you get a problem
that's more challenging. So let's prove this and
figure out why we can get away with ignoring this m g h right here. So let me get rid of all these. Let's start fresh. Let's just say we have a
spring hanging from the ceiling of spring constant k. Let's say this spring is light. If it was heavy, it might pull
itself down by its own weight so I'm gonna assume this
is very light spring and it's hanging right here. There's no mass connected to it initially so it's just hanging
at its natural length. It's neither pulling up nor pushing down as you see it right here because it's at its natural length but we connect the mass m to it. When we do that, we lower
the mass with our hand. We don't just let it fall
and start oscillating. We first lower the mass. We connect it and lower it. We find the point where the
mass is just gonna stay at rest. That would be the new
equilibrium position. So this right here is essentially our new equilibrium position. In other words, that's the
point where the net force on the mass would be zero. But this time that's not at
the spring's natural length. The way it was on the horizontal spring. This time the equilibrium
position is this place, the distance a away from
the spring's natural length because right now it's
battling the force of gravity. In other words the spring
force exerted upward k x minus the gravitational
force which is m times g has got to equal zero
in order for this mass to be an equilibrium. So we can actually figure
out what this distance a would have to be in
terms of given variables. Since at the equilibrium
position, x, the distance the spring has been stretched
has just gonna have to equal, m g divided by the spring constant k. This is what a is gonna equal. So the distance, the mass hangs down at the equilibrium position
from the natural length of the spring is just gonna be m g over k. This is a in this diagram. This is gonna be key. So we're gonna hold on to
this result right here. Well let's do this. Let's ask if we take this
mass and pull it down an extra amount b from the
new equilibrium position, well at that point the
forces won't be equal. The spring's gonna be stretched extra. It's gonna be pulling up with more force in the force of gravity. So this mass is gonna accelerate upward. It's gonna reach this equilibrium
position with some speed. It's gonna shoo pass that. Now the spring force is less
than the force of gravity and so gravity wins in that case and it keeps going back and forth. We're gonna ask ourselves the
same question we did before. If we pull this down, the distance b, what is the speed of the mass when it passes through
the equilibrium position? Again we're gonna use conservation
of energy to answer this. So we're gonna say that the
initial energy in our system is gonna equal to final
energy in our system. We're gonna choose two points. Let's choose initially
this point down here. We release the mass from rest when it's pulled down at distance b below the new equilibrium position. Then our final point
is gonna be right here at the equilibrium position because that's where we wanna
know the speed of the mass. So let's try to figure out
how much energy there is in the system initially if I pull this mass down and let it go. Well initially if I'm
just letting this mass go, the mass is starting from rest. If the mass starts from
rest, it's got no speed and if it's got no speed,
it's got no kinetic energy. So there's no kinetic energy to start with if this mass is starting from rest but there is gonna be
spring potential energy and there's gonna be a lot
of spring potential energy. Because think about it. Not only is this mass
stretching the spring passed the new equilibrium
position by an amount b but the new equilibrium
position itself is stretched from the spring's natural length a. This formula, when you have 1/2, the spring constant times the length that the spring has been stretched, that's the total amount the
spring has been stretched. So the total amount the
spring has been stretched from its natural length
is gonna be a plus b. I've got to square that whole term. This is how much spring
potential energy there's gonna be in the system initially. So how much gravitational potential energy we're gonna start with? Well, that's kind of hard to us because you can always
choose where you want your h equals zero reference line to be. In other words, I'm gonna
choose this lowest point here because that's often convenient
when choose this to be the h equals zero reference line. We'll measure all heights from that point. This is allowed because
it's only really differences in gravitational potential
energy that matter so you can do this. You just have to be
consistent with your choice. But with that choice where
this is h equals zero, the height my mass has
at the initial position is gonna be zero. So that means the
gravitational potential energy which is given by m g
h is also gonna be zero at that initial point. So in terms of initial
energies, this is all I've got. This is my total initial energy, just the energy from the spring. Now we can set that equal
to our final energies so we're gonna have any
spring potential energy here in our final position. You might think no
because the final position is at the new equilibrium position but remember this new equilibrium
position still displays from the spring's natural length. So what I have, I'm gonna have 1/2 k times the amount the spring has been stretched from its natural position. At this new equilibrium position, the amount that has been
stretched is just a. So I'm gonna write a here because that's how far
the spring has stretched at this new equilibrium position. I have to square that because that's 1/2 k x squared. We're gonna have kinetic energy. This mass is gonna gain
speed as it flies upward. It's gonna be moving with some speed when it gets up to that point. So the kinetic energy is
gonna be 1/2 times the mass, times the speed the mass has at the equilibrium position squared. That's what we want to determine. But there's also gonna be
gravitational potential energy. We said h is zero down here. So if the mass is not there, it's gonna have potential
energy due to gravity. If it's b above this point, look at it, we pulled it down b so when it gets back to
the equilibrium position, if this is h equals zero, that's gonna be h equals
b above where it started. So I have to put in m
times g times the b value. This length right here. Since in moving up this
mass gained m g times b a gravitational potential energy. So how do we make progress here? I wanna solve for the speed v but I've got this mess over here. Look, I've got a plus b squared so I better handle that first. So let's say we do the 1/2 k
and then we square this out. Remember we do FOIL. So it's First, Outer, Inner, Last. I'm gonna get a quantity of a squared and then this cross term, I'm get plus two times a times b and then plus b squared. That's what happens if I square
this whole term right here. These start looking really
bad but don't despair. Something great, something
wonderful is about to happen because I'm gonna set this
equal to the right hand side. So if we multiply out
on the left hand side, we're gonna get a 1/2
times k times a squared plus a 1/2 times k times this 2 a b term, plus I'm gonna get another one, 1/2 k times this b squared term. We can say that that's supposed to equal everything on the right hand side. So we can already see something
that we can cancel out. I've got a 1/2 k a squared on each side. So if we subtract that from both sides, I can get rid of that. This 1/2 here is gonna cancel this two and I'm left with k a
b on the left hand side plus 1/2 k b squared. But, what is k a? If you're clever and you look up here, you're like, wait a minute,
I remember what k a was. K times a, if we just
multiply both sides by k here, it's gotta equal m g because
that's what just the statement of equilibrium that at
the equilibrium position k times a has to equal m times g. So I can replace k times a
over on this left hand side with m times g. M g might be like, why
would I ever wanna do that? Why would I wanna replace k times a and I still multiply by this b? Why would I replace k a with m g? Because now that term is gonna
cancel with the other m g b on the other side of this equation. One half m b squared plus m g b but I've got m g b on the left and m g b on the right now. That's why I replace this k a with m g. I can subtract that from both sides and magically I just get
the exact same relationship we had for the horizontal spring measured from the new
equilibrium position. This is important so let me restate this. You can either when solving
these vertical spring problems, measure your spring
displacement from the natural unstretched length of the
spring like we did right here. We had to add a plus b because that was the distance from the natural unstretched
length of the spring all the way to where the mass was. You can do that and include
gravitational potential energy and get the right answer. But what we just saw is that
these terms always cancel so the alternative is
that you can just measure the spring extension,
the spring displacement from the new equilibrium position. If you do that, you just
leave off all mentioned of gravitational potential energy and you get the same answer. You can think of gravity
simply as shifting the equilibrium position down a bit and then the mass and spring behaving just as I would on a horizontal surface as long as you only think
about spring displacement from that new equilibrium position. So in other words, if
you were given a problem, let me get rid of all these. If you're just given a problem and you were told a three kilogram mass is hanging from a vertical spring of spring constant 50 newtons per meter and this line here represents
the equilibrium position, it's just hanging out right there at rest. You pull this mass down from the point where it was hanging at
rest, 0.3 meter and let go. You wanna figure out what
speed will it be going when it reaches equilibrium position. You can just do this. You can say, all right, down here it's got spring potential energy, 1/2 k is 50 newtons per meter times the amount the
spring has been stretched but I'm just gonna consider stretches from the new equilibrium position. So I'm just gonna do 0.3. I'm not gonna worry about the fact that the spring has actually
already been stretched to get to this equilibrium position. I'm just gonna put this
stretching from equilibrium. I'm gonna set that equal
to the kinetic energy the mass has at the equilibrium position. We know that m is three
kilograms times v squared. I'm not gonna include information about the gravitational
potential energy at all because I only measured the displacements from the new equilibrium position. So at this point, I can
just solve for my v. If you solve that algebraically for v, you cancel out the twos. You divide both sides by three and take a square root. You get to the speed of
the mass at equilibrium is gonna be 1.2 meters per second. So to recap, even though
it seems initially like vertical springs would be much harder than horizontal springs because you've got gravitational forces and
gravitational potential energy to worry about, if you measure
the spring displacement from the new equilibrium position as opposed to the natural spring length, you can simply use conservation of energy without mention of gravitational
potential energy at all.