Question

A force of 7.50N is applied to a spring whose spring constant is 0.289 N/cm. Find its change in length.​

Answer 1

Hook's law states that

$F=k\Delta x$

where F is the applied force, k is the spring constant, and $\Delta x$ is the change in length.

Plugging your values, we have

$7.5 = 0.289\Delta x \implies \Delta x = \dfrac{7.5}{0.289}\approx 25.95$