Answer to Question #171539 in Physics for Jerick

Question

An elastic cord is 61 cm long when a weight of 75 N hangs from it but is 85 cm long when a weight of 210 N hangs from it. What is the “spring” constant k of this elastic cord?

Answer 1

Let the length of the elastic cord at rest be $x$ meters. Then, according to the Hooke's Law we can write the spring constant $k$ for both cases:

$$k=\dfrac{F_1}{\Delta x_1}=\dfrac{75}{0.61-x},$$ $$k=\dfrac{F_2}{\Delta x_2}=\dfrac{210}{0.85-x}.$$

Then, we get:

$$\dfrac{75}{0.61-x}=\dfrac{210}{0.85-x},$$ $$75(0.85-x)=210(0.61-x),$$ $$135x=64.35,$$ $$x=\dfrac{64.35\ N\cdot m}{135\ N}=0.48\ m.$$

Therefore, we can calculate the spring constant $k$:

$$k=\dfrac{F_1}{\Delta x_1}=\dfrac{75\ N}{0.61\ m-0.48\ m}=577\ \dfrac{N}{m}.$$