Solution:
The energy stored in a spring is \(U = \frac{1}{2}kx^2\). If stretched by 1 cm, \(U_1 = 50\text{ J}\). If further stretched by 1 cm, total stretch becomes 2 cm, so the stored energy is \(U_2 = \frac{1}{2}k(2x)^2 = 4U_1 = 200\text{ J}\).
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
Leave a Reply