# Thread: Hooke's Law

1. ## Hooke's Law

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

$\displaystyle \int \frac{\lambda x}{l}$

but not sure about all of this.

if this is all correct would i do (0+0+0) = mgh + 1/2 mv^2 + $\displaystyle \int\frac{\lambda x}{l}$

then solve for v?

2. 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 $\displaystyle \tfrac12\lambda x^2$, where $\displaystyle \lambda$ is the modulus and x is the distance stretched (0.4m in this case).

3. Thanks for helping but i think i sorted it now

The potential energy in the string should be
do you mean the energy stored?

potential energy = $\displaystyle mgh$

energy stored = $\displaystyle \frac{1}{2} \frac{\lambda x^2}{l}$

looks more like what your getting at?