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Papessa [141]
3 years ago
10

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 ?
Physics
1 answer:
cupoosta [38]3 years ago
6 0
<h3>Answer:</h3>

\displaystyle U_s = 0.6 \ J

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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

<u>Physics</u>

<u>Energy</u>

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

  • U is energy (in J)
  • k is spring constant (in N/m)
  • Δx is displacement from equilibrium (in m)
<h3>Explanation:</h3>

<u>Step 1: Define</u>

k  = 7.50 N/m

Δx = 0.40 m

<u>Step 2: Find Potential Energy</u>

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