Question

A block of weight $W$ produces an extension of $9cm$ when it is hung by an elastic spring of length $60cm$ and is in equilibrium. The spring is cut into two parts, one of length $40cm$ and the other of length $20cm$. The same load $W$ hangs in equilibrium supported by both parts as shown in the figure. Find the extension (in $cm$ ) in the spring.

Answer 1

Answer: 2

Let $k$ be the spring constant of the original spring
$W=9k\Rightarrow k=\frac{W}{9}$
Spring constant is inversely proportional to the length of the spring.
Spring constant for the $40cm$ piece
$k_1=k\times \frac{60}{40}=\frac{3}{2}k$,
$k_2=k\times \frac{60}{20}=3k$
When connected in parallel, the equivalent spring constant
$k^{\prime }=k_1+k_2$
$\Rightarrow k^{\prime }=\frac{3}{2}k+3k=4\frac{1}{2}k$
Let $x$ ' be the new extension.
Then $k^{\prime }{x}^{\prime }=W=9k$
$\Rightarrow \frac{9}{2}k\times x^{\prime }=9k$
$\Rightarrow x^{\prime }=9k\times \frac{2}{9k}=2cm$