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

Let k be the spring constant of the original springW=9k⇒k=W9Spring constant is inversely proportional to the length of the spring.Spring constant for the 40 cm piece k1=k×6040=32k, k2=k×6020=3kWhen connected in parallel, the equivalent spring constantk′=k1+k2⇒k′=32k+3k=412kLet x' be the new extension. Then k′x′=W=9k⇒ 92k×x′=9k⇒x′=9k×29k=2cm