A spring of force constant $k$ is cut into two pieces such that one piece is twice as long as the other. Then the longer piece will have a force constant of:

3 k

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\frac{2 k}{3}

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\frac{3 k}{2}

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6 k

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Solution

The correct option is C $\frac{3 k}{2}$
Let Length $= L$
So new length $= \frac{L}{3}, \frac{2 L}{3}$
$K . L = F$ (any force)
$F = k^{'} \frac{2 L}{3}$

$k^{'} = \frac{3 K}{2}$

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A spring of force constant k is cut into two pieces such that one piece is twice as long as the other. Then the longer piece will have a force constant of:

The correct option is C 3k2Let Length =LSo new length=L3,2L3K.L=F(any force)F=k′2L3k′=3K2

A spring of force constant $k$ is cut into two pieces such that one piece is twice as long as the other. Then the longer piece will have a force constant of:

The correct option is C $\frac{3 k}{2}$
Let Length $= L$
So new length $= \frac{L}{3}, \frac{2 L}{3}$
$K . L = F$ (any force)
$F = k^{'} \frac{2 L}{3}$

$k^{'} = \frac{3 K}{2}$