A spring of force constant $15 ext{ }N/m$ is cut into two pieces. If the ratio of their length is $

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A spring of force constant $15	ext{ }N/m$ is cut into two pieces. If the ratio of their length is $1:3$, then the force constant of smaller piece is $\_\_\_\_$ $N/m$

Accepted Answer

$K\mathcal{l} =$ constant$${K\mathcal{l} = K'\left( \frac{\mathcal{l}}{4} ight) }{K' = 4	ext{ }K }{K' = 60	ext{ }N/m}$$

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