Answer to Question #95026 in Mechanics | Relativity for Darius Maharaj

Question

A spring has a natural length of 20cm. When a load of 50 N is applied, the spring becomes 22cm long in all. a) How great an extension does the 50 N force cause? b) If the spring obeys Hooke’s Law, what extension would you expect if a force of 150 N is applied to it? c) What will be the total length of the spring under that 150 N force?

Answer 1

According to Hooke's law, $F = k \Delta x = k |l_1 - l_0|$.

a) When the force $F_1 = 50 N$ is applied, the length of the spring is $l_1 = 22 cm$, so the extension is $\Delta_1 x= 22 cm - 20 cm = 2 cm$.

b) From the first case, the spring constant according to Hooke's law is $k = \frac{F_1}{\Delta_1 x}$ . The extension for the force $F_2 = 150 N$ is $\Delta_2 x = \frac{F_2}{k} = \frac{F_1}{F_0} \Delta_1 x = 6 cm$. The total length of the spring is $l_2 = l_0 + \Delta_2 x = 26 cm$.