Question

A watermelon is suspended from a spring and causes the spring to expand 0.54 m. If the spring constant is 131 N/m, what is the mass of the water melon?

Answer 1

We can find the mass of the watermelon from the Hooke's Law:

$$F=kx,$$ $$mg=kx,$$ $$m=\dfrac{kx}{g}=\dfrac{131\ \dfrac{N}{m}\cdot 0.54\ m}{9.8\ \dfrac{m}{s^2}}=7.2\ kg.$$

Answer:

$m=7.2\ kg.$