Question

A spring stretches 30 cm to hold up a 4 kg mass. How far would the spring stretch if it is used to accelerate a 9 kg mass upward at 5 m/s2.

Answer 1

The force required to accelerate 9 kilos to 5 m/s² is

$$F=ma.$$

According to Hooke's law, the spring needs to be stretched for

$$x=\frac Fk=\frac{ma}{k}.$$

The stiffness coefficient can be found from the experiment with 30 cm compression and 4 kg mass:

$$k=\frac{F_e}{x_e}=\frac{m_eg}{x_e}.\\\space\\ x=x_e\frac{ma}{m_eg}=0.3\frac{9\cdot5}{4\cdot9.8}=0.34\text{ m}.$$