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Minchanka [31]
3 years ago
10

20 J of energy was used to

Physics
1 answer:
Leto [7]3 years ago
6 0

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

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