Vertical Mass Spring System | Analyzing work

In summary, the "Vertical Mass Spring System" examines the dynamics of a mass attached to a spring that oscillates vertically under the influence of gravitational and spring forces. The analysis includes calculating the work done on the system, understanding potential and kinetic energy transformations, and applying Hooke's Law. It highlights the importance of equilibrium positions and the impact of displacement on energy changes, ultimately providing insights into harmonic motion and energy conservation in mechanical systems.
  • #1
I_Try_Math
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Homework Statement
A massless spring with force k = 200 N/m constant hangs from the ceiling. A 2.0-kg block is attached to the free end of the spring and released. If the block falls 17 cm before starting back upwards, how much work is done by friction during its descent?
Relevant Equations
F = kx
Is there a typo in this question? Supposing there was no friction, the block would fall until the force of the spring was equal to ##mg = 2 * 9.8 = 19.6##, taking the upward y direction as positive. Since ##F_{spring} = -200y## and ##19.6 = -200(-0.098)##, the block would fall 9.8 cm. It's not possible for the block to fall farther than that with friction opposing the motion?
 
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  • #2
The point when ##F_{\rm spring} = -mg## is not the turn around point. It is the point where the mass is no longer accelerating downwards, but instead starts to slow down. Your contradiction therefore rests on a false premise.
 
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  • #3
Orodruin said:
The point when ##F_{\rm spring} = -mg## is not the turn around point. It is the point where the mass is no longer accelerating downwards, but instead starts to slow down. Your contradiction therefore rests on a false premise.
Interesting, that makes sense. Thank you for pointing that out.
 
  • #4
Can you perhaps think of a different approach? Have you tried using energy methods?
 
  • #5
Orodruin said:
Can you perhaps think of a different approach? Have you tried using energy methods?
Well, my initial thoughts were to try to use ##ME_{final} = KE_{final} + PE_{final} + W_{friction} = 0 - 0 + W_{friction}##? But if that's correct it doesn't seem to get me anywhere.

Edit: However if I reconsider energy if there was no friction maybe I can solve it.
 
  • #6
Are you certain you have the right LHS? Also, how do you define your potential energies?
 
  • #7
Orodruin said:
Are you certain you have the right LHS? Also, how do you define your potential energies?
I'm not really certain if the LHS is correct, but at least in the case without friction ##ME_{initial} = KE_{initial} + PE_{initial} ##. And since ##ME_{initial} = 0, PE_{initial} = 0## and ##KE_{initial} = 0##. ##ME_{final} = ME_{initial}= 0 = KE_{final} + PE_{final} = 0 + PE_{final} ## which implies ##PE_{final} = 0?##

Potential energy, the way I tried to define it, decreases in the -y direction.
 
  • #8
I_Try_Math said:
Potential energy, the way I tried to define it, decreases in the -y direction.
Does it? Are you trying to account only for gravitational potential energy? That is not the only potential energy at work here!
 
  • #9
Orodruin said:
Does it? Are you trying to account only for gravitational potential energy? That is not the only potential energy at work here!
I'm trying to make sense of the net potential energy, but if my equations relating the initial and final mechanical energy are right, then I'm struggling to see how it could be anything other than zero. I guess what I'm saying is that in my mind the potential energies of spring and gravitation are "cancelling" out but that's incorrect?
 
  • #10
I_Try_Math said:
I'm trying to make sense of the net potential energy, but if my equations relating the initial and final mechanical energy are right, then I'm struggling to see how it could be anything other than zero.
You have not yet explained why you would think they are zero. What are the contributions to the potential energy? What are their mathematical expressions?
 
  • #11
Orodruin said:
You have not yet explained why you would think they are zero. What are the contributions to the potential energy? What are their mathematical expressions?
It's kind of counterintuitive, but I guess what I was trying to say was basically that the potential energies of gravitation and the spring would cancel out. I mean, in the case without friction, if ##ME_{final} = 0 = ME_{initial}## then it has to be that ##PE_{final} = 0## because ##KE_{final}=0##. But that's incorrect?
 
  • #12
Perhaps you might consider using the work-energy theorem that says that the sum of all the works done on the block is equal to the change in kinetic energy. $$\Delta K=\sum_i W_i.$$You know that, from start to finish, the change in kinetic energy is zero, so add up all the works and set them equal to zero.
 
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  • #13
I_Try_Math said:
. . . if ##ME_{final} = 0 = ME_{initial}## then . . .
That cannot happen. The final and initial mechanical energies cannot be equal because we are told that there is friction that eats up some of the mechanical energy as the block is descending.
 
  • #14
kuruman said:
That cannot happen. The final and initial mechanical energies cannot be equal because we are told that there is friction that eats up some of the mechanical energy as the block is descending.
Is it correct to say that in the case without friction, at the turn around point when the mass reaches it lowest position, the net potential energy is zero? I understand that it's maybe irrelevant to this case with friction but I'm still trying to see if I understand it correctly.
 
  • #15
I_Try_Math said:
Is it correct to say that in the case without friction, at the turn around point when the mass reaches it lowest position, the net potential energy is zero? I understand that it's maybe irrelevant to this case with friction but I'm still trying to see if I understand it correctly.
Gravitational potential energy of zero depends where you define the zero point. It's only the change in GPE that is relevant.
 
  • #16
PeroK said:
Gravitational potential energy of zero depends where you define the zero point. It's only the change in GPE that is relevant.
If I pick the zero of gravitational potential energy to be at the spring's equilibrium point, what would the net potential energy be at the lowest position of the mass if it was like this problem except without friction?
 
  • #17
I_Try_Math said:
If I pick the zero of gravitational potential energy to be at the spring's equilibrium point, what would the net potential energy be at the lowest position of the mass if it was like this problem except without friction?
That should be clear. If the KE is zero at both points, then all the GPE lost is converted to EPE.
 
  • #18
I am still missing some equations that are actually applied to the problem in question rather than just describing the general idea. I believe that if you start writing down the relevant equations then things will tend to materialise.
 
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  • #19
Orodruin said:
I am still missing some equations that are actually applied to the problem in question rather than just describing the general idea. I believe that if you start writing down the relevant equations then things will tend to materialise.
Thankfully, I was finally able to work through to the correct solution using energy considerations, as you said.
 

FAQ: Vertical Mass Spring System | Analyzing work

What is a vertical mass-spring system?

A vertical mass-spring system consists of a mass attached to a spring that is oriented vertically. The system can oscillate up and down due to the restoring force of the spring and the gravitational force acting on the mass.

How do you calculate the work done by the spring in a vertical mass-spring system?

The work done by the spring can be calculated using the formula \( W = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position. This formula assumes that the spring obeys Hooke's Law.

What role does gravitational potential energy play in a vertical mass-spring system?

In a vertical mass-spring system, gravitational potential energy must be considered in addition to the elastic potential energy of the spring. The total mechanical energy of the system is the sum of the kinetic energy, gravitational potential energy, and elastic potential energy.

How does the equilibrium position in a vertical mass-spring system differ from a horizontal one?

In a vertical mass-spring system, the equilibrium position is where the force of the spring balances the gravitational force acting on the mass. This position is lower than the natural length of the spring due to the weight of the mass. In a horizontal system, the equilibrium position is simply where the spring is neither stretched nor compressed.

How do damping and external forces affect the work done in a vertical mass-spring system?

Damping forces, such as friction or air resistance, dissipate energy from the system, reducing the amplitude of oscillations over time. External forces can add energy to the system or remove it, depending on their direction and magnitude. Both factors alter the total work done on the system and can affect its long-term behavior.

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