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

The correct option is A

0.1 joule

$△P.E=\frac{1}{2}k(x_2^2-x_1^2)=\frac{1}{2}\times 10\left[\right(0.25{)}^2-\left(0.20{)}^2\right]$

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