Question

What is the spring constant for a spring if pulling it back 0.391 m causes a force in the opposite direction of 0.922 N?0.361 N-0.361 N2.36 N-2.36 N

Answer 1

We are asked to determine the spring constant given that it is stretched a distance of 0.391 meters and there is a restoring force of 0.922 Newtons.

To do that we will use Hooke's law:

$F=-kx$

Where:

$\begin{gathered} F=\text{ restoring force} \\ k=\text{ spring constant} \\ x=\text{ deformation} \end{gathered}$

Now, we solve for the spring constant by dividing both sides by "-x":

$\frac{F}{-x}=k$

now, we plug in the values:

$\frac{-0.922N}{-0.391m}=k$

Now, we solve the operations:

$2.36N=k$

Therefore, the spring constant is 2.36 Newtons.