Maximum speed of object between 2 springs

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Homework Help Overview

The problem involves a 1.50 kg box moving between two springs on a frictionless surface, initially compressed against the stronger spring. The question focuses on determining the maximum speed of the box after it is released from rest.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy principles to relate kinetic and potential energy. There are questions about the correct setup of the energy equations and the assumptions regarding initial conditions, particularly the initial velocity.

Discussion Status

Some participants have provided guidance on the correct use of energy terms, suggesting that only one spring's potential energy should be considered at a time. There is an acknowledgment of a mistake in unit conversion, which has been identified as a critical factor in the calculations.

Contextual Notes

Participants are working under the assumption that the initial velocity is zero, which simplifies the energy equations. There is a focus on ensuring correct unit conversions and understanding the implications of energy conservation in the context of the problem.

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[SOLVED] Maximum speed of object between 2 springs

Homework Statement



A 1.50 kg box moves back and forth on a horizontal frictionless surface between two different springs, as shown in the accompanying figure. The box is initially pressed against the stronger spring, compressing it 4.00 cm, and then is released from rest.
(I solved part A, B is: What is the maximum speed the box will reach?)

Homework Equations


K[tex]_{i}[/tex] + EPE[tex]_{i}[/tex] = K[tex]_{f}[/tex] + EPE[tex]_{f}[/tex]

.5mv[tex]^{2}[/tex] [tex]_{i}[/tex] + .5k[tex]_{i}[/tex]x[tex]_{i}[/tex] [tex]^{2}[/tex] = .5mv[tex]^{2}[/tex][tex]_{f}[/tex] + .5k[tex]_{f}[/tex]x[tex]^{2}[/tex][tex]_{f}[/tex]

(sorry about the messed-up subscripts; I can't make it work right)

The Attempt at a Solution


I solved for x[tex]_{}f[/tex] first, obtaining the correct answer of 5.66 cm.

I then plugged in values for the other variables: m of 1.5 kg, initial velocity of 0 (because it started from rest?), initial k of 32 N/m, initial x of 4 cm, final k of 16 N/m, final x of 5.66 cm. I converted cm to m and tried to solve for final velocity, but I got .062 m/s and it's supposed to be 1.85 (I got it wrong enough times that it gave me the answer, I just don't know how to solve for it).
 
Last edited:
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I am not sure exactly how you are trying to solve for the maximum velocity. Can you show your calculations and reasoning? I can't help find your mistake if I can't see it.
 
v[tex]_{f2}[/tex] = mv[tex]_{i}[/tex][tex]^{2}[/tex] + k[tex]_{i}[/tex]x[tex]_{i}[/tex][tex]^{2}[/tex] - k[tex]_{f}[/tex]x[tex]_{f}[/tex][tex]^{2}[/tex]

I used the conservation of energy rule to set the initial and final kinetic energy and elastic potential energy equal to each other, then tried to solve for final kinetic energy by subtracting both sides by final elastic potential energy. Then I divided the final kinetic energy by the mass to get final velocity squared. On the other side, that left initial velocity squared (its mass canceled out when I divided by the mass from the other side) plus initial spring constant * inital compression, minus final spring constant * final compression
After I plugged in the numerical values stated above, I took the square root to change final velocity squared to just final velocity.
 
Last edited:
Your problem is that you have both the potential energy terms involved.

When the block has potential energy in one spring, the other spring is not compressed at all. So, the total energy is either:

[tex]TE=1/2k_ix_i^2[/tex] or [tex]TE=1/2k_fx_f^2[/tex]

Thus, only one of the above should be involved in the equation you posted above.
 
OH! I feel stupid now... lol. I'll try that and see if I can make my problem work! Thanks.
 
The assumption I was relying on was initial velocity is 0... since there were no values aside from initial compression that could be used to get velocity. Is this a correct assumption? Because with a zero velocity, initial kinetic energy is also 0, so that would leave me with final kinetic energy equal to initial elastic potential energy. So initial EPE divided by mass gives final v^2?

edit: oh! I got it right... my last fatal mistake was a unit conversion of cm to m. lol. But I made it! Thank you again for your help.
 
Last edited:
Anytime.:smile:
 

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