What is the significance of height in a vertical spring system?

[andrewasd]

Homework Statement

Consider a block that is connected to the surface of the Earth with a vertical spring.
You know that the mass of the block is 1.4 kg, the stiffness of the spring is 520.5 N/m,
the relaxed length of the spring is 0.7 m, the initial height of the block is 0.4 m, the final
height of the block is 1.2 m, and that the block is initially moving vertically downward
with a velocity of magnitude 9.9 m/s. What must be the magnitude of the block’s
velocity in the final state?

Relevant Equations

none

a

 

[haruspex]

Homework Statement:: Consider a block that is connected to the surface of the Earth with a vertical spring.
You know that the mass of the block is 1.4 kg, the stiffness of the spring is 520.5 N/m,
the relaxed length of the spring is 0.7 m, the initial height of the block is 0.4 m, the final
height of the block is 1.2 m, and that the block is initially moving vertically downward
with a velocity of magnitude 9.9 m/s. What must be the magnitude of the block’s
velocity in the final state?
Relevant Equations:: none

a

You surely know some relevant equations associated with springs, velocity and mechanical energy.
You must also show some attempt, per forum rules.

 

[andrewasd]

Yes sorry, I am new to the forums. I've attempted with Hookes Law, and have attempted it but am confused

 

[berkeman]

Please post your work so far. We need to see that before we can offer tutorial help.

 

[andrewasd]

I'm guessing that we use the equation F(_spy) = -k*Change in length. The change in length is 0.8m because that is just 1.2-0.4. I also know the magnitude equation, which is the sqrt(first velocity ^2 + second velocity^2). I then drew it out on a graph to help visualize

 

[haruspex]

I'm guessing that we use the equation F(_spy) = -k*Change in length. The change in length is 0.8m because that is just 1.2-0.4. I also know the magnitude equation, which is the sqrt(first velocity ^2 + second velocity^2). I then drew it out on a graph to help visualize

No, that equation is wrong. It is not the change in length between two arbitrary lengths; the magnitude of the force is k*(difference between current length and relaxed length). Neither 1.2m nor 0.4m is the relaxed length.

But forces are not the best way to approach this problem. What else do you know about springs that you can connect more directly with masses and velocities?

 

[andrewasd]

I'm confused? I thought the most straightfoward way to do it would be Hookes law, F=-k*x but you say not to use forces? Is there an equation that I can get the velocity from directly? Maybe it's v=sqrt((k/m)*change in length ^2)

 

[berkeman]

Hint: What equations do you know for the energies involved in this problem? There are several forms of energy involved...

 

[andrewasd]

I'm not sure what youre getting at sorry, I thought it was one of the two equations I've mentioned. I'm confused. is it the 1/2*m*v^2 ?

 

[berkeman]

So one way to work on vertical motion problems is to consider whether energy is conserved or not. If the process is not lossy (no friction), then generally the total energy stays the same during the motion of a body (as long as there is no energy input like a push or a rocket motor or something). So when solving for a problem where an object falls from a height (ignoring air resistance), the total energy TE = KE + PE is constant. KE is the kinetic energy that you mention in your post above, and PE is the gravitational potential energy which involves the mass of the object and its height. Do you know that equation?

And in this problem, you have the added influence of the spring. Do you know the equation for the potential energy stored in a spring based on its extension from the relaxed length?

 

[andrewasd]

Ok I think I'm starting to get it, the PE equation would be m*g*h I belive, and so the answer would the magnitude of TE. The equation your talking about it I think is U=1/2 k x^2, but am not completely sure, how do you then add it to the total energy though?

 

[haruspex]

Maybe it's v=sqrt((k/m)*change in length ^2)

That's the sort of thing, but again you need to be careful what you mean by "change in the length" of a spring.
Rather than memorise a formula like that, without apparently understanding it, it is much better to work with the fundamental concepts from which it is obtained. Let's unravel it a bit:
Squaring both sides to get rid of the square root:
$v^2=\frac km(\Delta x)^2$
Multiplying to get rid of the division:
$mv^2=k(\Delta x)^2$.
Can you now recognise what each side represents? Based on that, can you clarify what each variable represents and what conditions are assumed?

 

[berkeman]

Good. So the total energy of the system when it's isolated (after it has been extended or compressed and released) is a constant. So it looks something like this:
$$TE = KE + PE_g + PE_s$$
where the last two terms are the gravitational PE and the PE in the compressed/extended spring. Gravitational PE gets higher as the height of the mass increases. Can you fill in the equation now with the terms you've written, and be sure to get the sign on the gravitational PE correct with respect to the other energies?

 

[andrewasd]

Ok so in that case, the final equation would be the magnitude of KE(1/2 * 1.4 * 9.9 ^2) + PE_g(1.4 * 9.81 * 0.3) + PE_s(1/2 * 520.5 * 0.5^2) and the units would be m/s because its a velocity, is this correct?

 

[haruspex]

Ok so in that case, the final equation would be the magnitude of KE(1/2 * 1.4 * 9.9 ^2) + PE_g(1.4 * 9.81 * 0.3) + PE_s(1/2 * 520.5 * 0.5^2) and the units would be m/s because its a velocity, is this correct?

That's an expression, not an equation (no =).
Please explain how you get the 0.3 and 0.5 distances.

 

[andrewasd]

Right that's my mistake, it would = the final velocity, or v_f. I got 0.3 because I thought that the distance was the compressed distance, so since equilibrium is 0.7 minus compressed distance of 0.4 equals 0.3. And for the other one I thought it was the peak from equilibrium, so 1.2-0.7. Am I mistaken in those distances?

 

[haruspex]

it would = the final velocity

But each of those terms is an energy, so cannot equal a velocity.
What does it equal, according to post #13?

I got 0.3 because I thought that the distance was the compressed distance

You need the height of the mass above some baseline. You can choose the baseline to be whatever you like provided you are consistent. (We'll come that again later.) So choosing 0.7 is ok, but you want the height above it. Anyway, the problem statement gives you that it starts at 'height 0.4', so a baseline has been chosen for you. There is no benefit in choosing a different one.

the peak from equilibrium, so 1.2-0.7

Yes, 1.2-0.7 is the right value to use here, but because 0.7 is the relaxed length (=height) of the spring, not the equilibrium length (height). We don't know what the equilibrium length is, though we could compute it.

 

[andrewasd]

Ok the baseline distance makes sense. I thought it was the relaxed height minus the start height. But I'm not sure what it equals then? And if it doesn't equal the velocity then why do that equation at all?

 

[haruspex]

Ok the baseline distance makes sense. I thought it was the relaxed height minus the start height. But I'm not sure what it equals then? And if it doesn't equal the velocity then why do that equation at all?

In this problem, because the spring is vertical, the height of the mass at each time is significant in two ways. It tells you about the gravitational potential energy and about the spring potential energy.

For the purpose of GPE, all that matters is the change in height, so what you use for a baseline is arbitrary. This is because, in the biosphere, gravitational acceleration is near enough constant, making GPE proportional to height, so change in GPE is proportional to change in height.

But for SPE, the energy is as the square of the extension or compression from the relaxed length. So you cannot use an arbitrary base length here, you must use the difference between the current length and the relaxed length.

The problem setter has chosen to express the height of the mass relative to the ground level at which the spring is mounted. As a result, the spring extension equals the height minus the relaxed length. You might as well use this same height for GPE.

I hope that is clearer. Either way, just go ahead and write and post equations. Things should get clearer as we proceed.