Question

A block keeps compressed a spring whose elastic constant K = 205 N / m. How much is the potential energy when the spring is deformed by 0.25 m?

Help please

Answer 1

`\boxed {\boxed {\sf 6.40625 \ Joules}}`

Step-by-step explanation:

Elastic potential energy can be found using the following formula.

`U=(1)/(2) kx^2`

Where U is the elastic potential energy, k is the elasticity constant, and x is the extension/spring stretch length.

We know the elasticity constant is 205 N/m and the spring is deformed by 0.25 meters.

`k= 205 \ N/m \\x=0.25 \ m`

Substitute the values into the formula.

`U=(1)/(2)(205 \ N/m) *(0.25 \ m )^2`

Solve according to PEMDAS: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.

First, solve the exponent.

• (0.25 m)²= (0.25 m)(0.25 )=0.0625 m²

`U=(1)/(2) (205 \ N/m)(0.0625 \ m^2)`

Multiply 205 and 0.0625.

`U=(1)/(2)(12.8125 \ N/m)`

Multiply 12.8125 and 1/2 or divide 12.8125 by 2.

`U=6.40625 \ N/m`

• 1 Newton meter is equal to 1 Joule.
• Therefore, our answer of 6.40625 N/m is equal to 6.40625 J

`U= 6.40625 \ J`

The elastic potential energy is 6.40625 Joules.

Answer 2

PE = 6.40625 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

SI Unit: Newton per meter N/m = Joules J

Elastic Potential Energy: PE = 1/2k(Δx)²

• k is spring constant
• Δx is the displacement from equilibrium

Step-by-step explanation:

Step 1: Define

Spring Constant K = 205 N/m

Δx = 0.25 m

Step 2: Find Potential Energy

1. Substitute in variables: PE = 1/2(205 N/m)(0.25 m)²
2. Evaluate Exponents: PE = 1/2(205 N/m)(0.0625 m²)
3. Multiplication: PE = 1/2(12.8125 N/m)
4. Multiplication: PE = 6.40625 N/m

Our Potential Elastic Energy would be 6.40625 Joules.