Question

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

• A. $0.1$
  
• B. $0.2$
  
• C. $03$
  
• D. $0.5$
  

Answer 1

The correct option is A $0.1$

Initial PE $=0.5\times k\times x^2=0.5\times 10\times {0.2}^2=0.2J$

Final PE $=0.5\times k\times x^2=0.5\times 10\times {0.25}^2=0.3125J$

Increase in PE $=0.3125-0.2=0.1125\approx 0.1J$

Hence correct answer is option $A$