Question

1. What is the spring constant of a spring that stretches 12 cm when an object weighing 24 Nis hung from it?

Answer 1

ANSWER:

200 N/m

STEP-BY-STEP EXPLANATION:

Given:

x = 12 cm = 0.12 m

F = 24 N

The force is given by the following equation:

$F=kx$

We substitute and calculate for k, which is ls spring constant , just like this:

$\begin{gathered} 24=k\cdot0.12 \\ \\ k=\frac{24}{0.12} \\ \\ k=200\text{ N/m} \end{gathered}$

The spring constant is 200 N/m