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A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring? 0. 57 m 0. 64 m 0. 80 m 1. 25 m.

Answer :

Lanuel

The displacement of this spring is equal to 0.80 meter.

Given the following data:

  • Elastic potential energy = 5184 Joules
  • Spring constant = 16,200 N/m.

To determine the displacement of this spring:

Mathematically, the elastic potential energy is given by the formula;

[tex]U_s = \frac{1}{2} kx^2[/tex]

Where:

  • Us is the elastic energy of a spring.
  • k is the spring constant.
  • x is the displacement of a spring.

Making x the subject of formula, we have:

[tex]x=\sqrt{\frac{2U_s}{k} }[/tex]

Substituting the parameters into the formula, we have;

[tex]x=\sqrt{\frac{2 \times 5184}{16200} }\\\\x=\sqrt{\frac{10368}{16200} }\\\\x=\sqrt{0.64}[/tex]

x = 0.80 meter.

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