Solution:
Energy is \(U = \frac{1}{2} k x^2\). Since \(U \propto x^2\), doubling the stretch from \(1\text{ cm}\) to \(2\text{ cm}\) increases the stored energy by a factor of \(2^2 = 4\). Thus, \(U' = 4 \times 50\text{ J} = 200\text{ J}\).
When a spring is stretched by 1 cm, it stores energy 50 J. If it is further stretched by 1 cm, the stored energy will be
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