1. ## Hooke's law (easy?)

One end of a light elastic string, of natural length 0.8, is attached to a fixed point o, and the other end is attached to a particle of mass 5kg. When the particle hangs in equilibrium vertically below o, the length of the string is 1.3m

a) Calculate the modulus of elasticity of the string.

b) Determine the energy stored in the string.

so i've written down

m=5 l=0.8 x=0.5

i know

kinetic energy = $\displaystyle \frac{1}{2}mv^2$

and

Potential energy = $\displaystyle mgh$

I don't actually know what potential energy and kinetic energy is but i know those equations and i know i need to use them somehow!

ok so for kinetic energy i assume straight away is 0 as the particle isn't moving therefore v=0

for potential energy i get

5*9.8*0.5 = 24.5N

and i realise that as it hangs in equilibrium - forces up = forces down.

does this mean that T(tension in the string) = Potential energy?

therefore T=24.5?

I also know the modulus of elasticity is denoted by $\displaystyle \lambda$

and that Force = $\displaystyle \frac{\lambda x}{l}$

so would i be right to form this equation

$\displaystyle 24.5=\frac{\lambda*o.5}{0.8}$ and the solve for $\displaystyle \lambda$?

I also know that energy stored is $\displaystyle \frac{\lambda*x^2}{2l}$

so obviously i need to complete part a to solve this easy equation :/

2. Originally Posted by djmccabie
so would i be right to form this equation

$\displaystyle 24.5=\frac{\lambda \times 0.5}{0.8}$ and the solve for $\displaystyle \lambda$?

so obviously i need to complete part a to solve this easy equation :/
Yes, so what is the problem, solve for $\displaystyle \lambda$

CB

3. i dont know if i have done part a right

Im thinking i may be making it more complicated than it is. is it right to say T=mg

so T=5*9.8

?

i can resolve it from there

4. Originally Posted by djmccabie
i dont know if i have done part a right
Part a comes first because solving it makes the rest easier.

The equation:

$\displaystyle mg=\frac{\lambda x}{l_0}$

is just Hook's law.

CB