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

According to the Hook's law, the force acting on the spring is equal to:

$$F = kx$$

where $k = 131N/m$ is the spring constant, and $x = 0.54m$ is the expansion of the spring. On the other hand, the force acting on the spring from the water melon is basically its weight. Thus

$$F = mg$$

where $m$ is the mass of the water melon and $g = 9.81m/s^2$ is the gravitational acceleration. Substituting this expression into the Hook's law, obtain:

$$mg = kx\\ m = \dfrac{kx}{g} = \dfrac{131\cdot 0.54}{9.81} \approx 7.21kg$$

Answer. 7.21 kg.