A spring has length $l$ and force constant $K$. If is cut into two springs of length $l_1$ and $l_2$ such…
A spring has length $l$ and force constant $K$. If is cut into two springs of length $l_1$ and $l_2$ such that, $l_1=n l_2(n$ is integer). The force constant of the spring of length $l_2$ is
$\frac{(n+1) K}{n}$
$K$
$\frac{K}{(n+1)}$
$K(1+n)$
Solution
If $l$ is cut into $l_1$ and $l_2$, such that
$l_1+l_2=l$, where, $n l_2=l_1$
$\therefore l_2=\frac{l}{n+1}$
As we know from the elasticity relation $K=\frac{E A}{l}$, where $E$ is Young's modulus, $A$ area of cross-section and $l$ is the length of the spring.
$\therefore K_2=\frac{E A}{l_2}=\frac{(n+1) E A}{l}=(n+1) K$
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