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.

Question

Infinity Learn · Accepted Answer

Let k be the spring constant of the original spring

$W = 9 k \Rightarrow k = \frac{W}{9}$

Spring constant is inversely proportional to the length of the spring.

Spring constant for the 40 cm piece $k_{1} = k \times \frac{60}{40} = \frac{3}{2} k$ , $k_{2} = k \times \frac{60}{20} = 3 k$

When connected in parallel, the equivalent spring constant

$k^{'} = k_{1} + k_{2} \Rightarrow k^{'} = \frac{3}{2} k + 3 k = 4 \frac{1}{2} k$

Let x' be the new extension. Then $k^{'} x^{'} = W = 9 k$

$\Rightarrow \frac{9}{2} k \times x^{'} = 9 k \Rightarrow x^{'} = 9 k \times \frac{2}{9 k} = 2 cm$

Application of Newtons laws

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