Question

The elastic potential energy of a stretched spring is given by $E=50x^2$ where $x$ is the elongation of the spring in meter and $E$ is in joule, then the force constant of the spring is:

• A. $50Nm$
• B. $100N{m}^{-1}$
• C. $100N/m^2$
• D. $100Nm$

Answer 1

Answer: (B)

$100N{m}^{-1}$

Elastic potential energy is given by,
$E=\frac{1}{2}K{x}^{2}GivenE=50x^2$
Putting in above equation,

$\Rightarrow K=100N{m}^{-1}$