Question

A load of 49 N attached to a spring hanging
vertically stretches the spring 3.1 cm. The
spring is now placed horizontally on a table
and stretched 13 cm.
What force is required to stretch it by this
amount?
Answer in units of N.

Answer 1

Let's first find the spring constant from the Hookes law:

$$F = kx,$$ $$k = \dfrac{F}{x} = \dfrac{49N}{0.031m} = 1580 \dfrac{N}{m}.$$

Finally, we can find the force that is required to stretch the string by 13 cm:

$$F = kx = 1580 \dfrac{N}{m} \cdot 0.13 m = 205.4N.$$

Answer:

$F = kx = 205.4N.$