HELP Conservation of mechanical energy question

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Homework Help Overview

The discussion revolves around a conservation of mechanical energy problem involving a glider attached to a spring on a frictionless air track. Participants are tasked with calculating the speed of the glider at specific points during its motion after being released from a compressed state.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation of energy principles, questioning the use of gravitational potential energy versus spring potential energy in their calculations. Some express confusion about deriving velocity from energy equations.

Discussion Status

There is an ongoing exploration of the correct energy equations to use, with some participants providing guidance on the transformation of spring potential energy to kinetic energy. Multiple interpretations of the problem are being discussed, but no consensus has been reached.

Contextual Notes

Participants are working under the assumption that the system is isolated, and there is a focus on the energy transformations occurring as the spring is compressed and released. Some participants express uncertainty about their calculations and the correct application of formulas.

abot
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I am having problems with this question. I keep getting an answer of 1.87m/s but the answer is 1.47m/s.

A glider of mass .150kg moves on a horizontal frictionless air track. It is permanently attached to one end of a massless horizontal spring, which has a force of constant of 10 N/m both for extension and compression. The other end of the spring is fixed. The glider is moved to compress the spring by .180m and then released from rest. Calculate the speed of the glider (a) at the point where it has moved .180m from its starting point, so that the spring is momentarily exerting no force and (b) at the point where it has moved .250m from its starting point.
 
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abot said:
I am having problems with this question. I keep getting an answer of 1.87m/s but the answer is 1.47m/s.
Can't tell you where you went wrong if you don't show your work.
 
Wnc = K + U
0= .5(mv)^2 + mgh
v= sq.root{(2-(mgh)) / (m))
v= sq.root{2-(.15*9.8) / (.150)
v= sq.root(3.53)
v=1.87

I may be totally wrong but that's how i though it should be done.
 
You are you using gravitational PE--but you should be using spring PE ([itex]1/2 k x^2[/itex]).

As the spring stretches, spring potential energy is transformed to kinetic energy. The air track is horizontal, so gravitational PE doesn't change.
 
yes i understand why you would use such equation but how would you get velecity from it
 
[tex]E_m_i = E_p + E_k[/tex]
[tex]E_m_i = \frac{kx^2}{2} + 0[/tex]

Plug and chug.

Do the same thing for [tex]E_m_f[/tex]

Recall, this is an isolated system, therefore [tex]E_m_i = E_m_f[/tex]

Solve for v
 
alright... got it...thanks
 

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