Question

A rubber band stores 1.59 J of PEewhen it is stretched 0.292 m.What is its spring constant?(Unit=N/m)

Answer 1

Given:

The potential energy stored in the spring is P.E. = 1.59 J

The spring is stretched by x = 0.292 m

To find the spring constant.

Explanation:

The spring constant can be calculated by the formula

$\begin{gathered} P.E.\text{ = }\frac{1}{2}kx^2 \\ k=\frac{2\times P.E.}{x^2} \end{gathered}$

On substituting the values, the spring constant will be

$\begin{gathered} k\text{ =}\frac{2\times1.59\text{ J}}{(0.292)^2m^2} \\ =37.3\text{ N/m} \end{gathered}$

Thus, the spring constant is 37.3 N/m