Two springs of force constant 100 N/m and 150 N/m are in series as shown. The block is pulled by a distance of 2.5 cm to the right from equilibrium position. What is the ratio of work done on the spring at left to the work done on the spring at right?

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$\frac{3}{2}$

$\frac{2}{3}$

0.2

0.5

answer is A.

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Detailed Solution

$\begin{array}{l} W_{l e f t s p r i n g} = \frac{1}{2} k_{1} e_{1}^{2} = \frac{F^{2}}{2 k_{1}} \\ W_{r i g h t s p r i n g} = \frac{1}{2} k_{2} e_{2}^{2} = \frac{F^{2}}{2 K_{2}} \\ \frac{W_{l e f t}}{W_{r i g h t}} = \frac{k_{2}}{k_{1}} = \frac{150}{100} = \frac{3}{2} \end{array}$

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