1. Tardigrade
  2. Question
  3. Physics
  4. A massless spring of spring constant k , has extension y and potential energy E . It is now stretched from y to 2y . The increase in its potential energy is

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A massless spring of spring constant $ k $ , has extension $ y $ and potential energy $ E $ . It is now stretched from $ y $ to $ 2y $ . The increase in its potential energy is

J & K CETJ & K CET 2016Work, Energy and Power

Solution:

The potential energy of the spring is
$E=\frac{1}{2} ky^{2} \dots(i)$
Now it is stretched from $y$ to $2y$, so its potential energy becomes
$E'=\frac{1}{2}k(2y)^{2}=2ky^{2}=4E$ (using (i))
$\therefore $ The increase in its potential energy is
$\Delta \, E=E' -E$
$=4E-E=3E$