Question

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

• A. $\frac{1}{2}k(x_2-x_1{)}^2$
• B. $\frac{1}{2}k(x_2+2x_1)$
• 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. $\frac{1}{2}k(x_1^2+x_2{)}^2$

Answer 1

Answer: (B)

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