- Home
- Standard 11
- Physics
An elastic string of unstretched length $L$ and force constant $k$ is stretched by a small length $x$. It is further stretched by another small length $y$. The work done in the second stretching is
$\frac{1}{2} Ky ^2$
$\frac{1}{2}Ky(2x+y)$
$\frac{1}{2}K(x^2+y^2)$
$\frac{1}{2} k(x+y)^2$
Solution
(b)
In the string elastic force is conservative in nature.
$\therefore W =-\Delta U$
Work done by elastic force of string,
$W =-\left( U _{ F }- U _{ i }\right)= U _{ i }- U _{ F }$
$W = \frac{1}{2} kx ^2-\frac{ k }{2}( x + y )^2$
$=\frac{1}{2} kx ^2-\frac{1}{2} k \left( x ^2+ y ^2+2 xy \right)$
$=\frac{1}{2} kx ^2-\frac{1}{2} ky ^2-\frac{1}{2} kx ^2-\frac{1}{2} k (2 xy )$
$=- kxy -\frac{1}{2} ky ^2$
Therefore, the work done against elastic force
$W _{\text {external }}=- W =\frac{ ky }{2}(2 x + y )$
