A spring of force constant k is cut into two | vedclass

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Question

(C) Let the original spring have length $\ell_0$ and force constant $k$. The spring constant is inversely proportional to the length,i.e.,$k \propto 1

Vedclass · Accepted Answer

greater for the shorter part

Vedclass · Answer

(C) Let the original spring have length $\ell_0$ and force constant $k$. The spring constant is inversely proportional to the length,i.e.,$k \propto 1/\ell$.When the spring is cut into two parts of lengths $\ell_1 = \ell_0/3$ and $\ell_2 = 2\ell_0/3$,their new spring constants are:$k_1 = k \cdot (\ell_0 / \ell_1) = k \cdot (\ell_0 / (\ell_0/3)) = 3k$$k_2 = k \cdot (\ell_0 / \ell_2) = k \cdot (\ell_0 / (2\ell_0/3)) = 1.5k = 3k/2$Work done in stretching a spring by an amount $x$ is given by $W = \frac{1}{2} k x^2$.Since both parts are stretched by the same amount $x$,the work done is directly proportional to the spring constant $(W \propto k)$.For the shorter part (length $\ell_0/3$),$W_1 = \frac{1}{2} (3k) x^2 = 1.5 k x^2$.For the longer part (length $2\ell_0/3$),$W_2 = \frac{1}{2} (1.5k) x^2 = 0.75 k x^2$.Since $W_1 > W_2$,the work done is greater for the shorter part.

$A$ spring of force constant $k$ is cut into two parts at one-third of its length. When both parts are stretched by the same amount,the work done in the two parts will be:

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