QNEET-2023-Ph-232: NEET 2023 Physics - Class 11 NEET-2023-Ph-232 Solution: The potential energy of a long spring when stretched by 2 cm is …

The potential energy (U) stored in a spring when it is stretched or compressed is given by the formula: U = 1 2 k x 2 where: k is the spring constant, x is the displacement from the spring's equilibrium position (i.e., how much the spring is stretched or compressed). If the spring is initially stretched by 2 cm (or 0.02 meters, since we generally use SI units for these calculations), the potential energy stored can be represented as: U = 1 2 k ( 0.02 ) 2 When the spring is stretched by 8 cm (or 0.08 meters), the new potential energy U _ becomes: U _ = 1 2 k ( 0.08 ) 2 To find the relation between U _ and U , we can divide the expression for U _ by that of U : U _ U = 1 2 k ( 0.08 ) 2 1 2 k ( 0.02 ) 2 Simplifying this quotient gives: U _ U = ( 0.08 0.02 ) 2 = ( 4 ) 2 = 16 Thus, if the spring's displacement is increased from 2 cm to 8 cm, the potential energy stored in the spring increases by a factor of 16 , meaning: U _ = 16 U