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 J
• B. 0.2 J
• C. 0.3 J
• D. 0.5 J

Answer 1

The correct option is A 0.1 J

Potential energy of object at initial stretch

$P_1=\frac{1}{2}k\times 2=\frac{1}{2}\times 10\times 0.2\times 0.2=0.2J$

Potential energy of object at changed stretch

$P_2=\frac{1}{2}k\times 2=\frac{1}{2}\times 10\times 0.25\times 0.25=0.3J$

difference in potential energy = $P_2-P_1=0.3J-0.2J=0.1J$