**1. The problem statement, all variables and given/known data**

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

**2. Relevant equations**

E.P.E= λx^2/2l, v^2=u^2+2gh, P.E= mgh

**3. 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

http://ift.tt/1hMhdUT