Saved Bookmarks
| 1. |
The work done in stretching a spring of constant K by amount by delta |
|
Answer» Potential energy in a stretched spring U=21kx2 where, X is the EXTENSION in the spring. So, U2=21kl22 and U1=21kl12 Work DONE W=U2−U1 W=21kl22−21kl12=21k(l22−l12). |
|