Question

A block is attached to a spring having equilibrium position at $x=0$. The block oscillates between $x=-4\text{m}$ and $x=4\text{m}$. Find the change in potential energy stored in the spring when the block moves from $x=2$ to $x=-3$. Given the spring constant is $5\text{N/m}$.

• A. $12.5\text{J}$
• B. $32.5\text{J}$
• C. $37.5\text{J}$
• D. $10\text{J}$

Answer 1

The correct option is A $12.5\text{J}$

Energy stored when spring is at $x=2\text{m}$ is
$U_1=\frac{1}{2}k{x}^2=\frac{1}{2}\times 5\times 2^2$
$U_1=10\text{J}$

Energy stored when spring is at $x=-3\text{m}$ is
$U_2=\frac{1}{2}k{x}^2=\frac{1}{2}\times 5\times (-3{)}^2$
$U_2=22.5\text{J}$

Change in potential energy of the spring $=U_2-U_1$
$=22.5-10=12.5\text{J}$