Hooke’s Law

The formulas related to Hooke’s law are:

F = $-k * x$
$k = -\dfrac{-F}{\lambda}$
k = $\dfrac{(m*g)}{x}$

Example:

A block is suspended from the ceiling and is 12 cm from the ceiling. The block has a mass of 3 kg. The spring has an unstretched length of 2 cm. Find the spring constant and the amount of elastic potential energy stored in the spring when the block is at this height.

Given: $m = 3kg\quad g = 9.80 \quad x = 10cm$

Find: $k = ? \quad PE_{SP}=?$

We can use Hooke’s Law to find the spring constant and potential energy.

$k = [\frac{(3)(9.80)}{(0.10)}](\frac{1N}{(kgm/s^2) }= 294 N/m$

$PE_{SP}=\frac{1}{2}*k*x^{2}=\frac{1}{2}*(294N/m)*(0.10m)^{2}=(1.47N*m)*(\frac{1J}{1N*m})=1.47J$

$So:\quad k=294 N/m \quad PE_{SP}=1.47 J$