## Series of Springs

Consider three springs in series are attached to a fixed point on a ceiling as shown in the figure. If a 5 kg mass is attached at the end, how much will the springs stretch in total?

##
__
__**Hint**

**Hint**

The force in a spring is:

$$$F_s=k\delta $$$

where
$$k$$
is the spring constant, and
$$\delta$$
is the change in spring length from the un-deformed spring length.

##
__
__**Hint 2**

**Hint 2**

The spring constant for springs in series:

$$$\frac{1}{k_{eq}}=\sum_{i}\frac{1}{k_i}$$$

First, let’s determine the equivalent spring constant for several springs in series:

$$$\frac{1}{k_{eq}}=\sum_{i}\frac{1}{k_i}$$$

Since the problem statement has three springs in series:

$$$\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}=\frac{1}{4N/m}+\frac{1}{2N/m}+\frac{1}{3N/m}$$$

$$$\frac{1}{k_{eq}}=\frac{3}{12N/m}+\frac{6}{12N/m}+\frac{4}{12N/m}=\frac{13m}{12N}$$$

$$$\frac{1}{\frac{1}{k_{eq}}}=\frac{1}{\frac{13m}{12N}}$$$

$$$k_{eq}=\frac{12}{13}\:N/m$$$

A spring’s deflection and force are related by:

$$$F_s=k_{eq}\delta $$$

where
$$F_s$$
is the forced applied to the spring,
$$k_{eq}$$
is the equivalent spring constant, and
$$\delta$$
is the change in spring length from the un-deformed spring length.

Remember,
$$Force = mass \times acceleration$$
. Therefore, the spring will stretch a total of:

$$$\delta=\frac{F_s}{k_{eq}}=\frac{13m}{12N}\times (5kg)(9.8m/s^2)$$$

$$$=\frac{13m}{12N}\times 49N=53.1\:m$$$

53.1 m