Question

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 ?

Answer 1

Answer:

$\displaystyle U_s = 0.6 \ J$

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: $\displaystyle U_s = \frac{1}{2} k \triangle x^2$

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

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]:                                        $\displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.40 \ m)^2$
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:                                                                                                           $\displaystyle U_s = 0.6 \ J$