Answer:
[tex]K_{system} = \frac{k}{10}[/tex]
Explanation:
When the springs are connected end to end, it means they are connected in series. When the springs are connected in series, the stress applied to the system gets applied to each of the springs without any change in magnitude while the strain of the system is the sum total of strains of each spring. The spring constant of the resultant system is given as,
[tex]\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{n}})[/tex]
Here, n = 10
Spring constant of each spring = k
Thus,
[tex]\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{10}})[/tex]
[tex]\frac{1}{K_{system}} = (\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})[/tex]
[tex]\frac{1}{K_{system}} = \frac{10}{k}[/tex]
[tex]K_{system} = \frac{k}{10}[/tex]
Answer:
[tex]K_{eq}= \frac{K}{10}[/tex]
Explanation:
When springs are connected end to end their effective spring constant
[tex]\frac{1}{K_{eq}}=\frac{1}{K_1} +\frac{1}{K_2}.................................\frac{1}{K_{10}}[/tex]
all ten springs same spring constant K
⇒K=K1=K2........K10
therefore,
[tex]\frac{1}{K_{eq}}=\frac{1}{K} +\frac{1}{K}.................................\frac{1}{K}[/tex]
[tex]\frac{1}{K_{eq}} =\frac{10}{K}[/tex]
⇒[tex]K_{eq}= \frac{K}{10}[/tex]
