Answer to Question #275394 in Physics for nica

Question

A force of 486N stretches a spring by 0.08m. What is the force constant of the spring? What force will elongate the spring by 0.09m? What is the potential elastic energy when stretched by 0.18m?

Answer 1

The force constant:

$$k=F/x=6075\text { N/m}.$$

The force for 0.09 m:

$$F=kx_2=547\text{ N}.$$

The energy:

$$E=\frac12kx_3^2=98.4\text{ J}.$$