Saved Bookmarks
| 1. |
A spring of constant 100 N/m is stretched by applying equal forces each of magnitude Fat the two ends. The energy stored in the spring is 200 J. Now spring is cut into two equal parts and one of the part is stretched by applying equal forces each of magnitude F at the two ends. The energy stored is |
|
Answer» 200 J `therefore` Elastic potential energy stored in the spring is, `U=(1)/(2) kx_(0)^(2)` or `200= (F^(2) )/( 2K)` If spring is CUT into two parts, spring constant of each part becomes twice. `therefore F=2k x_(0) rArr x_(0) = (F)/( 2k)` `therefore` The energy stored is, `U.= (1)/(2) (2k) x_(0)^(2) = k(F^2)/(4k^(2))` `= (F^(2))/(4k)= (200)/(2) = 100` J. |
|