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].(adsbygoogle = window.adsbygoogle || []).push({});

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 cant seem to get the anwer using these, are they correct?

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# Homework Help: Mechanics - Elastic springs and strings

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