## Physics: Principles with Applications (7th Edition)

a. Here, the force is the same. $F = k_{1}x_{1} = k_{2}x_{2}$. The work done on spring 1 is $$W_{1} = \frac{1}{2} k_{1}x_{1}^{2}$$ The work on spring 2 is $$W_{2} = \frac{1}{2} k_{2}x_{2}^{2} = \frac{1}{2} k_{2}\frac{k_{1}^{2} x_{1}^{2}}{ k_{2}^{2}}$$ $$W_{2} = W_{1} \frac{k_{1}}{k_{2}}$$ Because the first spring is stiffer, the work done on spring 2 is greater. b. Now, the distance stretched is the same. As before, the work done on spring 1 is $$W_{1} = \frac{1}{2} k_{1}x^{2}$$ The work on spring 2 is $$W_{2} = \frac{1}{2} k_{2}x^{2}$$ Because the first spring is stiffer, the work done on spring 1 is greater.