Is Conservation of Energy Demonstrated in Particle Motion from A to P?

[Dumbledore211]

Homework Statement

A particle P of mass m is attached to one end of a light 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 5l/3. Show that if P is projected vertically downwards from A with speed √(3gl/2). P will come to instantaneous rest after moving a distance of 10l/3

Homework Equations

E.P.E= λx^2/2l, v^2=u^2+2gh, P.E= mgh

The Attempt at a Solution

Here, I have been asked to show that the total distance traveled by the particle is 10l/3 which we know is true if we think logically about it because when I take the particle up at A it travels 5l/3 and when it comes back to it's equilibrium position it again traverses 5l/3. When we lift the particle the work done in this case is P.E=mgh and when we project the particle vertically downwards then the work converts P.E into K.E and E.P.E where the equation of velocity is v^2=u^2-2gh → 0=u^2-2gh→ u^2=2gh. Should the total equation for the conservation of energy look like this that describes both the particle's action of going up and coming down
P.E= E.P.E + K.E

 

[voko]

I do not understand you explanation. The particle is not "taken up at A" and it does not travel 5l/3. The problem says that the entire length of the spring is 5l/3 in equilibrium, which a different statement.

Anyway, all you need here is to compute the total mechanical energy initially and the total mechanical energy at max elongation.

 

[Dumbledore211]

Shouldn't the equation of conservation of mechanical energy look like this P.E= E.P.E + K.E
mgh=λx^2/2l + mv^2/2

 

[Dumbledore211]

When I equate the whole thing like mgh=λx^2/2l + mv^2/2 taking velocity=√(3gl/2) I don't get h= 10l/3 why??

 

[voko]

I am not sure what λx^2/2l means. Is it (λx^2)/(2l) or is it (λx^2/2)l? And what is λ anyway? This does not look like stiffness in Hooke's law.