Question

A spring with force constant $k$ is initially stretched by $x_1$. If it further stretched by $x_2$, then the increase in its potential energy is

• A. $\frac{1}{2}k(x_2-x_1{)}^2$
• B. $\frac{1}{2}k{x}_2(x_2+2x_1)$
• C. $\frac{1}{2}k{x}_1^2+\frac{1}{2}k{x}_2^2$
• D. $\frac{1}{2}k(x_1+x_2{)}^2$
• E. $\frac{1}{2}k(x_1+x_2{)}^3$

Answer 1

The correct option is B $\frac{1}{2}k{x}_2(x_2+2x_1)$

According to the question, in the first condition
$U_1=\frac{1}{2}k{x}_1^2....\left(i\right)$
In the second condition,
$U_2=\frac{1}{2}k(x_1+x_2{)}^2....(ii)$
Increment in the potential energy
$=U_2-U_1$
$=\frac{1}{2}k(x_1^2+x_2^2+2x_1{x}_2)-\frac{1}{2}k{x}_1^2$
$=\frac{1}{2}k(x_1^2+x_2^2+2x_1{x}_2-x_1^2)$
$=\frac{1}{2}k[2x_1{x}_2+x_2^2]$
$=\frac{1}{2}k{x}_2(x_1+2x_1)$.