Question

A spring of force constant $10$ N/m has an initial stretch $0.20$ m. In changing the stretch to $0.25$ m, the increase in potential energy is about

• A. $0.1$ joule
• B. $0.2$ joule
• C. $0.3$ joule
• D. $0.5$ joule

Answer 1

Answer: (A)

$0.1$ joule

Change in potential Energy is given by

$\Delta P.E.=\frac{1}{2}k(x_2^2-x_1^2)$

$=\frac{1}{2}\times 10\left[\right(0.025{)}^2-\left(0.20{)}^2\right]$

$=5\times 0.45\times 0.05=0.1J$