Question

A spring of spring constant 100 N m–1 is expanded by 1 cm. The potential energy stored in it will be

Answer 1

• k = 100 N/m
• x = 1cm = 0.01 m

⇒U = ½kx²

⇒U = ½ * 100 * (0.01)²

⇒U = ½ * 100 * 0.0001

⇒U = ½ * 0.01

⇒U = 0.005 Joules.

Answer 2

Answer:

the potential energy stored in the spring is = 0.005 Joules

Explanation: -

                        Given: -  

             spring constant(k) = 100N/m

              expansion in the spring (x) = 1cm = 0.01 m

                  To find out: -

              the potential energy stored in the spring (U)=?

               FORMULA USED

                       U = 1/2kx²

               putting the given values in the desired equation

                       U = 1/2 × 100 × 0.01 × 0.01

                        U = 0.005 Joules

to learn more about spring constant kindly refer to link below: -

https://brainly.in/question/1603466

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