How to calculate the potential energy of a spring?

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SUMMARY

The discussion centers on calculating the potential energy (PE) of a spring-mass system. The correct formula for the potential energy is established as PE(x) = mgx + 1/2*k*(x/sin(a) - L)^2 + 1/2*k*(x/sin(a) - L)^2. Participants clarify that the gravitational potential energy (mgx) should be included only if the motion is vertical; otherwise, it should be omitted. The kinetic energy (KE) is defined as KE(x) = 1/2*m*(dx/dt)^2, emphasizing that the mass m is the only contributor to kinetic energy, not the springs.

PREREQUISITES
  • Understanding of potential energy in mechanical systems
  • Familiarity with Hooke's Law and spring constants (k)
  • Knowledge of gravitational potential energy (mgx)
  • Basic calculus for relating kinetic energy to velocity (dx/dt)
NEXT STEPS
  • Study the derivation of potential energy in spring systems
  • Learn about the principles of energy conservation in mechanical systems
  • Explore the application of differential equations in motion analysis
  • Investigate the effects of angle on spring force components
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, engineers working with spring systems, and anyone interested in energy calculations in mechanical contexts.

  • #31
haruspex said:
No, that's your job.
To recap:
In post #16 you had the right expression for the potential energy in the springs.

Unless you have some good reason to suppose it is in the vertical plane, go with Sammy's suggestion that is is horizontal, so discard any gravitational term.

In post #21 your force equation is quite wrong. You cannot add forces acting in different directions as though they are scalars. If you want the net force from the springs then you need to consider the x and y components of the forces. (This is the path berkeman was going down, but he wrongly tried to apply that to the PE of the springs instead of to their tensions.). But I don't think you need to worry about forces anyway, just concentrate on energy.

In post #21, your KE expression had two copies of 1/2 mv2. There is one mass m moving at speed v, not two.

From all that, see if you can put together the correct expression for the total energy.
the kinetic energy in function of x is : KE(x)= 1/2*m*v^2= 1/2*m*(dx/dt)^2
Is it correct now?
 
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  • #32
ramialsaiad said:
the kinetic energy in function of x is : KE(x)= 1/2*m*v^2= 1/2*m*(dx/dt)^2
Is it correct now?
Yes. So what is the total energy?
 
  • #33
haruspex said:
Yes. So what is the total energy?

total energi = 1/2*m*(dx/dt)^2 + mgx +1/2*k*(x/sin(a) - L)^2 + 1/2*k*(x/sin(a) - L)^2
the kinetic energy is 1/2*m*(dx/dt)^2 ... Can we rewrite the kinetic energy equation in function of x instead of dx ?
 
  • #34
ramialsaiad said:
total energi = 1/2*m*(dx/dt)^2 + mgx +1/2*k*(x/sin(a) - L)^2 + 1/2*k*(x/sin(a) - L)^2
the kinetic energy is 1/2*m*(dx/dt)^2 ... Can we rewrite the kinetic energy equation in function of x instead of dx ?
No, you cannot express it purely in terms of x except by solving the differential equation.
What is the whole question? You originally said you wanted to find the potential energy, which you have done. Why is the KE of interest? Do you need to find the equation of motion?

Also, I see you have reintroduced mgx. That is applicable if the motion is in the vertical plane, but not if it is in the horizontal plane. Do you know which plane it is in? If the question does not state, I would assume it is horizontal, and throw away the mgx term.
 
  • #35
haruspex said:
No, you cannot express it purely in terms of x except by solving the differential equation.
What is the whole question? You originally said you wanted to find the potential energy, which you have done. Why is the KE of interest? Do you need to find the equation of motion?

Also, I see you have reintroduced mgx. That is applicable if the motion is in the vertical plane, but not if it is in the horizontal plane. Do you know which plane it is in? If the question does not state, I would assume it is horizontal, and throw away the mgx term.
Why horizontal though?

The forces in the horizontal direction cancel each other ( If we assume that both of the springs have a constant of k) and the net force in the y direction. So It will only move vertically...?
 
  • #36
Biker said:
Why horizontal though?

The forces in the horizontal direction cancel each other ( If we assume that both of the springs have a constant of k) and the net force in the y direction. So It will only move vertically...?
Horizontal plane. That is, the diagram may be as viewed from above.
 
  • #37
haruspex said:
No, you cannot express it purely in terms of x except by solving the differential equation.
What is the whole question? You originally said you wanted to find the potential energy, which you have done. Why is the KE of interest? Do you need to find the equation of motion?

Also, I see you have reintroduced mgx. That is applicable if the motion is in the vertical plane, but not if it is in the horizontal plane. Do you know which plane it is in? If the question does not state, I would assume it is horizontal, and throw away the mgx term.
I have to find KE , it is required ..
I do want to find the equation of motion in terms of x for KE and PE and the force on the brick as a function of x ..
Iam sure it is moving in vertical direction ..
 
  • #38
ramialsaiad said:
I have to find KE , it is required ..
I do want to find the equation of motion in terms of x for KE and PE and the force on the brick as a function of x ..
Iam sure it is moving in vertical direction ..
Ok, if you are sure it is vertical keep the mgx.
What type of motion do you think it will be?
 
  • #39
haruspex said:
Ok, if you are sure it is vertical keep the mgx.
What type of motion do you think it will be?
potential energy and I got it correct as you said before
 
  • #40
ramialsaiad said:
potential energy and I got it correct as you said before
Not sure what question you are answering. I asked what kind of motion you thought would result.
 
  • #41
haruspex said:
Not sure what question you are answering. I asked what kind of motion you thought would result.
no idea
 
  • #42
ramialsaiad said:
no idea
In the simpler case of a mass hanging from a spring, or lying on a smooth table and connected to a horizontal spring, what type of motion occurs?
 
  • #43
haruspex said:
In the simpler case of a mass hanging from a spring, or lying on a smooth table and connected to a horizontal spring, what type of motion occurs?
potential energy
 
  • #44
ramialsaiad said:
potential energy
That is not a type of motion. Here are some types: constant velocity, constant acceleration, elliptical orbit, simple harmonic motion. Pick one.
 
  • #45
haruspex said:
That is not a type of motion. Here are some types: constant velocity, constant acceleration, elliptical orbit, simple harmonic motion. Pick one.
simple harmonic motion
 
  • #46
ramialsaiad said:
simple harmonic motion
Right. Can you write the generic form of the equation for that? If so, try that out as a solution to your differential equation.
 
  • #47
haruspex said:
Right. Can you write the generic form of the equation for that? If so, try that out as a solution to your differential equation.
I do not know how to do ?
if you can provide the starting equation to me that will be nice ,, then I try to go further
 
  • #48
ramialsaiad said:
I do not know how to do ?
if you can provide the starting equation to me that will be nice ,, then I try to go further
I find it very strange that you have been given a problem in harmonic motion yet appear to know so little about it.
The general form of the solution is ##x=A\sin(\omega t+\phi)##, where A is the amplitude, ω is the frequency and φ is the phase. Or you can use cos instead of sin, which changes the value of φ but keeps all else the same.
 

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