Finding Velocity in Spring Problem

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    Spring Velocity
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Homework Help Overview

The discussion revolves around a physics problem involving the conservation of energy in a spring system, specifically focusing on finding the velocity of an object after it has been released from a certain height. The problem incorporates concepts of gravitational potential energy, spring potential energy, and kinetic energy.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the energy conservation equation and question whether the initial conditions and variables have been correctly accounted for, particularly regarding the height and spring compression.

Discussion Status

Some participants have offered insights into the potential miscalculation regarding the height used in the energy equation. There is an acknowledgment of the need to incorporate additional factors, such as spring compression, into the calculations. The discussion is ongoing, with attempts to clarify the setup and identify errors.

Contextual Notes

There appears to be confusion regarding the correct application of the energy conservation equation, particularly in how the initial height and spring compression are defined and utilized in the calculations.

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Homework Statement



314ealu.jpg


Homework Equations



K1 + Ug1 + Us1 = K2 + Ug2 + Us2

The Attempt at a Solution



Need to find h1, and that is h1 = Dsin(theta) = .92sin(39) = .579 m

.5mv1^2 + mgh1 + .5kx^2 = .5mv2^2 + mgh2 + .5kx^2

0 + mgh1 + .5kx^2 = .5mv2^2 + 0 + 0

1.7*9.8*.579 + .5*170*.18^2 = .5*1.7v2^2

12.40014 = .85v2^2

v2 = 3.8195 m/s

My answer is wrong apparently, not quite sure what I did wrong either as it seems all the work is right. I guess I set it up wrong, is it supposed to be mgh1 = .5mv2^2 + .5kx^2 ?
 
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Need to find h1, and that is h1 = Dsin(theta) = .92sin(39) = .579 m


to find h1 use L = (0.92 + 0.18)
 
rl.bhat said:

Need to find h1, and that is h1 = Dsin(theta) = .92sin(39) = .579 m


to find h1 use L = (0.92 + 0.18)

Ah I see now, I was just using the .92 and not utilizing the compression. I'll try it again and see if I get it correct. Is that my only error then?
 
I think so.
 

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