Question

When a spring is compressed 2.50 × 10^–2 meter
from its equilibrium position, the total potential
energy stored in the spring is 1.25 × 10^–2 joule.
Calculate the spring constant of the spring.
[Show all work, including the equation and
substitution with units.]

Answer 1

     The working of a spring is given by:

$T_{el}= \frac{k\Delta x^2}{2}$
 
     Entering the unknowns, we have:

$T_{el}=\frac{k\Delta x^2}{2} \\ 1.25*10^{-2}*2*J=k(2.5*10^{-2}m)^2 \\ k= \frac{2.5*10^{-2}*J}{(2.5*10^{-2}m)^2} \\ k=40* \frac{J}{m^2} \\ k=40 \frac{N*m}{m^2} \\ \boxed {k=40* \frac{N}{m} }$
 
If you notice any mistake in my english, please let me know, because i am not native.