Answer:
[tex]k_{eq} = \frac{k}{N}[/tex]
Explanation:
For this exercise let's use hooke's law
F = - k x
where x is the displacement from the equilibrium position.
x = [tex]- \frac{F}{k}[/tex]
if we have several springs in series, the total displacement is the sum of the displacement for each spring, F the external force applied to the springs
x_ {total} = ∑ x_i
we substitute
x_ {total} = ∑ -F / ki
F / k_ {eq} = -F [tex]\sum \frac{1}{k_i}[/tex]
[tex]\frac{1}{k_{eq}} = \frac{1}{k_i}[/tex] 1 / k_ {eq} = ∑ 1 / k_i
if all the springs are the same
k_i = k
[tex]\frac{1}{k_{eq}} = \frac{1}{k} \sum 1 \\[/tex]
[tex]\frac{1}{k_{eq} } = \frac{N}{k}[/tex]
[tex]k_{eq} = \frac{k}{N}[/tex]
