Question

The work done by external agent in stretching a spring of force constant $k=100N/cm$ from deformation $x_1=10$ to deformation $x_2=20cm$.

• A. $-150J$
• B. $50J$
• C. $150J$
• D. None of these

Answer 1

Answer: (C)

$150J$

Work done by external agent =change in potential energy of spring

$=\frac{1}{2}k{x}_2^2-\frac{1}{2}k{x}_1^2$

$=\frac{1}{2}\left(10000N/m\right)(0.2{)}^2-\frac{1}{2}(10000\left)\right(0.1{)}^2$

$=5000(0.04-0.01)=150J$