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

PE = 6.40625 J

Pre-Algebra

Order of Operations: BPEMDAS

Energy

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

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

Step 1: Define

Spring Constant K = 205 N/m

Δx = 0.25 m

Step 2: Find Potential Energy

Our Potential Elastic Energy would be 6.40625 Joules.

Explanation:

Elastic potential energy can be found using the following formula.

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.

Substitute the values into the formula.

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

First, solve the exponent.

Multiply 205 and 0.0625.

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

The elastic potential energy is 6.40625 Joules.