Question

You start with spring that's already been stretched an unknown amount from equilibrium. After stretching it an additional 2.0 cm, you find that the total potential energy associated with the spring is 18 J. You stretch it another 2.0 cm, and now the potential energy rises to 25 J. Find the spring constant.

Answer 1

Answer: 35*10^3 N/m

Explanation: In order to explain this problem we know that the potential energy for spring is given by:

Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.

We  also know that with additional streching of 2 cm of teh spring,  the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.

Then the difference of energy for both cases is 7 J so:

ΔUp= 1/2*k* (0.02)^2 then

k=2*7/(0.02)^2=35000 N/m