Answer to Question #305240 in Physics for Bliss

Question

A spring 20cm long is stretched to 25cm by a load of 50N. What will be it's length when stretched by 100N assuming that the elastic limit is not reached

Answer 1

The spring constant can be determined from the Hook's law:

$$k = \dfrac{F_1}{\Delta l_1} = \dfrac{50N}{25cm - 20cm} =10N/cm$$

Then, the expansion under the second force:

$$\Delta l_2 = \dfrac{F_2}{k} = \dfrac{100N}{10N/cm} =10cm$$

Thus, the length of the spring will be:

$$l_2 = 20cm+10cm = 30cm$$

Answer. 30 cm.