That approach looks correct. h should be –1.2 metres, because the particle P has dropped that distance. The potential energy in the string should be , where is the modulus and x is the distance stretched (0.4m in this case).
A light elastic string, of natural length 0.8m and modulus of elasticity 35.4N, has one end A attached to a fixed point and the other end B attached to particle P of mass 3kg. Initially P is held at rest at A. It is then released and allowed to fall. Calculate the speed of P when the length of the string is 1.2m
OK this is how i've been taught to answer the question.
energy at start = potential energy + kinetic energy + energy stored
= 0 + 0 + 0
energy at greatest extension = potential energy + kinetic energy + energy stored
P.E. = mgh
[I dont know what h is]
K.E. = 1/2 mv^2
i believe energy stored is 'Hooke's Law' integrated
but not sure about all of this.
if this is all correct would i do (0+0+0) = mgh + 1/2 mv^2 +
then solve for v?