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1) A spring, which has a spring constant k=7.50 N/m, has been stretched 0.40 m from ts equilibrium position . What the potential energy now tored in the spring ?

Respuesta :

Space

Answer:

[tex]\displaystyle U_s = 0.6 \ J[/tex]

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Physics

Energy

Elastic Potential Energy: [tex]\displaystyle U_s = \frac{1}{2} k \triangle x^2[/tex]

  • U is energy (in J)
  • k is spring constant (in N/m)
  • Δx is displacement from equilibrium (in m)

Explanation:

Step 1: Define

k  = 7.50 N/m

Δx = 0.40 m

Step 2: Find Potential Energy

  1. Substitute in variables [Elastic Potential Energy]:                                        [tex]\displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.40 \ m)^2[/tex]
  2. Evaluate exponents:                                                                                      [tex]\displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.16 \ m^2)[/tex]
  3. Multiply:                                                                                                           [tex]\displaystyle U_s = (3.75 \ N/m) (0.16 \ m^2)[/tex]
  4. Multiply:                                                                                                           [tex]\displaystyle U_s = 0.6 \ J[/tex]

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