Question

a spring contains 1.6 J of elastic potential energy when it is extended by 0.40m. what is it's spring constant?

Answer 1

The elastic potential energy of a spring is given by the expression
`PE = (1)/(2) kx^2`, where k is the spring constant and x is the extension of spring.

So we have
`k = (2PE)/(x^2)`

PE = 1.6 Joule

x = 0.40 meter

Substituting

`k = (2*1.6)/(0.4^2) \\ \\k =20 N/m`

So spring constant K = 20 N/m