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When a spring is stretched by 2 cm its potential energy?

Sagot :

The spring constant is 50000 N/m.

The spring constant is a measure of the stiffness of a spring.

  • Springs with higher stiffness are more difficult to stretch.
  • Under the action of external forces, the spring is deformed and returns to its original position when the force is removed.
  • Spring constant is denoted by 'k' and has a SI unit as N/m.
  • It has different values for different springs and different material.

Electric potential energy of a spring can be calculated as P.E. = (1/2)*k*x²

where, k is the spring constant,

x is the displacement caused by the spring

The spring stores energy = 100 J

Displacement suffered by the spring = 2 cm = 0.02 m

P.E. = (1/2)*k*x²

k = (2*P.E.) / x²

k =  (2 * 100) / (0.02)²

k = 200 / 0.0004

k = 50000 N/m

The spring constant for a spring having potential energy as 100 J  is 50000 N/m.

The question is incomplete the complete question is "When a spring is stretched by 2 cm its potential energy is 100J? What is the spring constant".

To know more about spring constant,

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