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?

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  = k e²

k is the spring constant

e is the extension

  Insert the parameters and solve;

          20  = x 445 x e²

        multiply those sides by 2;

         2(20) = 445e²

            40  = 445e²

              e² =   = 0.09

              e  = √0.09 = 0.3m

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2