Answer to Question #198124 in Physics for Eren Yeager

Question

Two spring with the same force constant of 75 N/m are connected end to end, and the

combination is pulled out by a force of 20 N. a) What is the PE of the stretched springs?

b) If the two springs are connected side-by-side and pulled by the same force, what will be it’s

PE now?

Answer 1

By definition, PE of the stretched springs is given as follows:

$$PE = \dfrac{kx^2}{2}$$

where $k$ is the total force constant of two springs, and $x$ is their elongation.

a) In case of series connection the total spring constant will be the following (see https://socratic.org/questions/what-is-the-spring-constant-in-parallel-connection-and-series-connection):

$$k = \dfrac{k_1 k_2}{k_1+k_2} = \dfrac{k_1}{2}$$

where $k_1 = k_2 = 70N/m$ are the spring constants of two springs.

The elongation is given by the Hook's law:

$$x = \dfrac{F}{k} = \dfrac{2F}{k_1}$$

where $F = 20N$ is the applied force. Thus, obtain:

$$PE = \dfrac{k_1\cdot (2F/k_1)^2}{2\cdot 2} = \dfrac{F^2}{k_1}\\ PE = \dfrac{20^2}{70} \approx 5.7J$$

b) For the springs in parallel, have:

$$k = k_1 + k_2 = 2k_1\\ x = \dfrac{F}{k} = \dfrac{F}{2k_1}\\ PE = \dfrac{2k_1\cdot (F/2k_1)^2}{2} = \dfrac{F^2}{4k_1}\\ PE = \dfrac{20^2}{4\cdot 70} \approx 1.4J$$

Answer. a) 5.7 J, b) 1.4 J.