Find the max compression of the spring

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Homework Help Overview

The discussion revolves around finding the maximum compression of a spring in a system involving kinetic energy (KE), gravitational potential energy (GPE), and spring potential energy (SPE). Participants are analyzing energy conservation principles and the relationships between different forms of energy in the context of a mechanical system.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the conservation of energy by setting initial and final energy equations equal to each other, questioning the inclusion of GPE and the treatment of velocities for connected bodies.

Discussion Status

Some participants have offered guidance on correcting signs in equations and clarifying the treatment of energy types. There is ongoing exploration of different interpretations of the energy equations, with no explicit consensus reached on the correct approach.

Contextual Notes

Participants note confusion regarding the initial conditions of the system, the treatment of negative values in calculations, and the implications of energy types on the final results. There is also mention of homework constraints that may limit the exploration of certain methods.

isukatphysics69
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Homework Statement



help.PNG

Homework Equations


ke = .5mv^2

The Attempt at a Solution


Ei= .5m1v21 + .5m2v22 + PEs(PEs initially 0)

Ef= .5m1v21 + .5m2v22 + PEs(PEs is now .5(16.2)(.2672)set them equal and then solve for velocity right? is this the right approach?
 

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isukatphysics69 said:

Homework Statement



View attachment 225073

Homework Equations


ke = .5mv^2

The Attempt at a Solution


Ei= .5m1v21 + .5m2v22 + PEs(PEs initially 0)

Ef= .5m1v21 + .5m2v22 + PEs(PEs is now .5(16.2)(.2672)set them equal and then solve for velocity right? is this the right approach?
What about GPE?
 
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Why do you use two different velocities for two tied bodies?
And the system is set in motion by the gravity, and there is no GPE in the equations.
 
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omg thank you guys completely forgot gpe also yes i know the velocities are the same but just labeled like that thank you
@haruspex @Sergio Rodriguez
 
ok what is going on here i took into account gpe and got 1.59m/s and incorrect
GPE = mgdcosθ
initially 0 because distance was 0 at initial state
The energy of the whole system is initially 0
 
wait a second i will try something here i see something wrong. spe is negative, the spring wants to snap back i was using positive
 
ok i need help
initial energy is 0 for this system, final energy is
EFINAL = .5m1v2FINAL + .5m2v2FINAL +.5(16.2)(.2672)+((.481)(9.8)(-.267))
So

0 = .5m1v2FINAL - .5m2v2FINAL +.5(16.2)(.2672)+((.481)(9.8)(-.267))
solve for vFINAL
why is this not correct?
 
isukatphysics69 said:
ok i need help
initial energy is 0 for this system, final energy is
EFINAL = .5m1v2FINAL + .5m2v2FINAL +.5(16.2)(.2672)+((.481)(9.8)(-.267))
So

0 = .5m1v2FINAL - .5m2v2FINAL +.5(16.2)(.2672)+((.481)(9.8)(-.267))
solve for vFINAL
why is this not correct?
Why the sign change on the second term?
 
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haruspex said:
Why the sign change on the second term?
sorry that is not what i meant i meant

0 = .5m1v2FINAL + .5m2v2FINAL -.5(16.2)(.2672)+((.481)(9.8)(-.267))

i meant to put negative spring potential energy sorry for confusion
 
  • #10
ok i just got the answer by doing something i was telling myself not to do
so it says speed. I was getting a negative value and i was going to square root it but knew i didn't want to square root a negative. since it says speed i decided to omit the negative and got the correct answer of 1.37m/s
 
  • #11
0 = .5m1v2FINAL - .5m2v2FINAL - .5(16.2)(.2672)+((.481)(9.8)(.267))

and solving for v gave the correct answer

couple of things here, i thought that the ((.481)(9.8)(.267)) at the end of the equation should have been ((.481)(9.8)(-.267)) because it was moving downward
and when i was doing the algebra i got a negative value under the square root. i had to multiply it by negative 1 to get a positive and then took the square root and got the correct answer. My gut was telling me not to do that for hours and i have tried this way earlier and saw a negative under the square root and instantly gave up and tried a different approach.
 
  • #12
isukatphysics69 said:
i meant to put negative spring potential energy
Why? The KE and spring PE are both gains in energy. The only energy lost is the GPE.
isukatphysics69 said:
ok i just got the answer by doing something i was telling myself not to do
so it says speed. I was getting a negative value and i was going to square root it but knew i didn't want to square root a negative. since it says speed i decided to omit the negative and got the correct answer of 1.37m/s
Your reasoning was correct. v2 should not have turned out negative.
I am mystified as to how you got the right answer from that equation.
Go back to your post #7, correct the sign on the m2v2 term and try again.
 
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  • #13
isukatphysics69 said:
0 = .5m1v2FINAL - .5m2v2FINAL - .5(16.2)(.2672)+((.481)(9.8)(.267))

and solving for v gave the correct answer

Yes, the KE is always positive.
The GPE could be zero at the lowest level and positive at the initial position or begin with zero and negative later. But the SPE is positive because the spring is stretching and gaining energy.
 
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