Question

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

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

Answer 1

The correct option is C $100J/m^2$

Let the force constant of the spring be $k$.

Elastic potential energy of a spring is given by $E$ =$\frac{1}{2}k{x}^2$. ....(i)

Given , elastic potential energy of stretched spring is $E$ = $50$ $x^2$ ....(ii)

Comparing equations (i) and (ii) :

$\frac{1}{2}k{x}^2$ = $50$ $x^2$

We get $k$ = $100$ $\frac{J}{m^2}$ .