Question

An elastic spring of unstretched length $L$ and force constant $K$, is stretched by a small length $x$. It is further stretched by another small length $y$. calculate the work done during the second stretching is:

• A. $\frac{ky}{2}(x+2y)$
• B. $\frac{k}{2}(2x+y)$
• C. $ky(x+2y)$
• D. $\frac{ky}{2}(2x+y)$

Answer 1

Answer: (D)

$\frac{ky}{2}(2x+y)$

Initial stretching of the spring is $x$.
Initial potential energy $U_i=\frac{1}{2}k{x}^2$
Final stretching of the spring is $x+y$.
Final potential energy of the spring $U_f=\frac{1}{2}k(x+y{)}^2=$ $\frac{1}{2}k{x}^2+\frac{1}{2}k{y}^2+\frac{1}{2}k\left(2xy\right)$
Work done $W=U_f-U_i=$ $\frac{1}{2}k{x}^2+\frac{1}{2}k{y}^2+\frac{1}{2}k\left(2xy\right)-\frac{1}{2}k{x}^2$
$⟹W=\frac{ky}{2}(2x+y)$