InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The potential energy of a spring, when stretched through a distance x is 50J. What would be the work done in stretchin it further through the same distance? |
|
Answer» Here, `x_(1=)x, E_(1)=(1)/(2)kx^(2)=50J, W=?x_(2)=x+x=2x` As work done `=` increase in P.E. of the spring `:.W=E_(2)-E_(1)=(1)/(2)kx_(2)^(2)-(1)/(2)kx_(1)^(2)=(1)/(2)k[(2x)^(2)-x^(2)]W=3xx(1)/(2)kx^(2)=3xx50=150J` |
|