A long spring is stretched by 2 ~cm and its potential energy is U . If the spring is stretched by 10 ~cm ; its potential energy will be

Step-by-step solution

The potential energy of a stretched spring is U= 1 2 k x² Here, k= spring constant, x= elongation in spring. But given that, the elongation is 2 ~cm . So, U= 1 2 k(2)² U= 1 2 k 4 ...(i) If elongation is 10 ~cm then potential energy aligned U^ &= 1 2 k(10)² U^ &= 1 2 k 100 ...(ii) aligned On dividing Eq. (ii) by Eq. (i), we have U^ U = 1 2 k 100 1 2 k 4 or U^ U =25 U^ =25 U