Question

A spring of force constant 10 N/m has initial stretch 0.2 m. In changing the stretch to 0.25 m, the increase of PE 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

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

Here

$x_1=0.20cm{x}_2=0.25cmD.P.E=\frac{1}{2}\times 10\left(\right(0.25{)}^2-\left(0.20{)}^2\right)D.P.E=5\times 0.45\times 0.05P.E=0.1J$