A long spring, when stretched by a distance x , has potential energy U . On increasing the stretching to n x , the potential energy of the spring will be

Step-by-step solution

The potential energy U stored in a spring stretched by a distance x is given by the formula U = 1 2 k x² , where k is the spring constant. When the stretching is increased to n x , the new potential energy U' is given by U' = 1 2 k (n x)² . Expanding the expression, we get U' = 1 2 k n² x² . Substituting the original potential energy U = 1 2 k x² into the expression for U' , we obtain U' = n² ( 1 2 k x²) = n² U . Answer: n^2 U