Mechanics - Elastic springs and strings

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SUMMARY

The discussion centers on the mechanics of a particle P attached to an elastic string, demonstrating that when projected downwards with a speed of \(\sqrt{\frac{3gl}{2}}\), it will come to instantaneous rest after moving a distance of \(\frac{10l}{3}\). The participant outlines the energy calculations before and after projection, noting the kinetic energy (KE) as \(\frac{3gl}{4}\) and gravitational potential energy (GPE) as \(-(y - \frac{5}{3}L)g\). The elastic potential energy (EPE) is expressed as \(\frac{gl(Y - \frac{5}{3}L)^2}{2L(Y - \frac{5}{3}L)}\). The participant seeks confirmation on the correctness of these energy equations.

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  • Understanding of elastic potential energy (EPE)
  • Knowledge of gravitational potential energy (GPE)
  • Familiarity with kinetic energy (KE) concepts
  • Basic principles of differential equations
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mr bob
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A particle P of mass m is attached to one end of a light elastic string of natural length L whose other end is attached to a point A on a ceiling. When P hangs in equilibrium AP has length [itex]\frac{5l}{3}[/itex]. Show that if P is projected vertically downwards from A with speed [itex]\sqrt(\frac{3gl}{2})[/itex], P will come to instantaneous rest after moving a distance [itex]\frac{10l}{3}[/itex].

I thought about working all this out by finding the energies before and after the projection.

Before:-
[itex]KE =\frac{3gl}{4}[/itex]
[itex]GPE = 0[/itex]
[itex]EPE = 0[/itex]

After:-
[itex]KE = 0[/itex]
[itex]GPE = -(y - 5/3L)g[/itex] where y is the full length of stretched string.
[itex]EPE = \frac{gl(Y- 5/3L)^2}{2L(Y- 5/3L)}[/itex]

i can't seem to get the anwer using these, are they correct?
 
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This seems like more of a physics related question. I'm sure if you post it in the physics section someone would help you out a little faster. I haven't taken differential equation in awhile, but this seems like a pretty straightforward differential equation question. I could definitely be wrong though
 

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