Question

20 J of energy was used to
stretch a spring with a
spring constant of 445 N/m,
what is the final extension
of the spring?

Answer 1

Answer:

0.3m

Explanation:

Given parameters;

Elastic energy  = 20J

Spring constant = 445N/m

Unknown

Final extension of the spring  = ?

Solution:

The elastic potential energy of a stretched spring can be determined using the expression below;

         

           EP  = $\frac{1}{2}$ k e²

k is the spring constant

e is the extension

  Insert the parameters and solve;

          20  = $\frac{1}{2}$ x 445 x e²

        multiply those sides by 2;

         2(20) = 445e²

            40  = 445e²

              e² = $\frac{40}{445}$   = 0.09

              e  = √0.09 = 0.3m