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

0.3m

Step-by-step 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 =
`(1)/(2)` k e²

k is the spring constant

e is the extension

Insert the parameters and solve;

20 =
`(1)/(2)` x 445 x e²

multiply those sides by 2;

2(20) = 445e²

40 = 445e²

e² =
`(40)/(445)` = 0.09

e = √0.09 = 0.3m