Question

A spring has a total length of 17.5 cm under a load of 250 g and 20.0cm under a load of 300 g. The extension of the spring per unit load is

Answer 1

According to Hooke's law:

$$F=k\Delta x=k(x_e-x_0),$$

where $\Delta x$ is elongation, $x_e$ is the length of extended spring, $x_0$ is the initial length. Hence, we can write two equations:

$$m_1g=k(x_{e1}-x_0),\\ m_2g=k(x_{e2}-x_0).\\ 0.25\cdot9.8=k(0.175-x_0),\\ 0.3\cdot9.8=k(0.2-x_0).\\ k=19.6\text{ N/m},\\ x_0=5\text{ cm}.$$

The extension of the spring per unit load is, therefore

$$\kappa=\frac{1}{k}=0.051\text{ m/N}.$$