**Elastic potential energy** (U_{s}) is half times the product of **spring constant** (k) and the square of **displacement** (Δx). Using the formula of elastic potential energy: **U _{s}**

**= ½ × k Δx**, the value of elastic potential energy of spring can be calculated. (Δx is

^{2}**+ve**when spring is stretched and Δx is

**-ve**when spring is compressed)

Let’s solve some problems based on this formula, so you’ll get a clear idea.

## Elastic Potential Energy Practice Problems

**Problem 1:** Calculate the elastic potential energy of a spring when it is stretched from 0.15 m to 0.30 m. (Take the value of spring constant, k = 150 N/m)

Solution:

Given data:

Elastic potential energy of a spring, U_{s} = ?

Initial length of a spring, x_{i} = 0.15 m

Finial length of a spring, x_{f} = 0.30 m

Displacement of spring, Δx = x_{f} – x_{i} = 0.15 m

Spring constant, k = 150 N/m

Using the formula of elastic potential energy,

U_{s} = ½ × k Δx^{2}

U_{s} = ½ × 150 × (0.15)^{2}

U_{s} = ½ × 150 × 0.0225

U_{s} = 1.68 J

Therefore, the elastic potential energy of a spring is **1.68 J**.

**Problem 2:** A spring having an initial length of 5 m is compressed to a final length of 2 m. Calculate the elastic potential energy of a spring. (Take the value of spring constant, k = 120 N/m)

Solution:

Given data:

Initial length of a spring, x_{i} = 5 m

Finial length of a spring, x_{f} = 2 m

Displacement of spring, Δx = x_{f} – x_{i} = -3 m

Elastic potential energy of a spring, U_{s} = ?

Spring constant, k = 120 N/m

Using the formula of elastic potential energy,

U_{s} = ½ × k Δx^{2}

U_{s} = ½ × 120 × (-3)^{2}

U_{s} = ½ × 120 × 9

U_{s} = 540 J

Therefore, the elastic potential energy of a spring is **540 J**.

**Problem 3:** A spring is stretched from its rest length of 0.25 m to 0.45 m with a spring constant k = 180 N/m. Calculate the elastic potential energy stored in a spring.

Solution:

Given data:

Initial length of a spring, x_{i} = 0.25 m

Finial length of a spring, x_{f} = 0.45 m

Displacement of spring, Δx = x_{f} – x_{i} = 0.2 m

Spring constant, k = 180 N/m

Elastic potential energy of a spring, U_{s} = ?

Using the formula of elastic potential energy,

U_{s} = ½ × k Δx^{2}

U_{s} = ½ × 180 × (0.2)^{2}

U_{s} = 90 × 0.04

U_{s} = 3.6 J

Therefore, the elastic potential energy stored in a spring is **3.6 J**.

**Problem 4:** A 0.85 m spring is fixed in the wall. Calculate the elastic potential energy stored in a spring, when it is compressed to a length of 0.35 m. (Take the value of spring constant, k = 130 N/m)

Solution:

Given data:

Initial length of a spring, x_{i} = 0.85 m

Finial length of a spring, x_{f} = 0.35 m

Displacement of spring, Δx = x_{f} – x_{i} = -0.5 m

Elastic potential energy of a spring, U_{s} = ?

Spring constant, k = 130 N/m

Using the formula of elastic potential energy,

U_{s} = ½ × k Δx^{2}

U_{s} = ½ × 130 × (-0.5)^{2}

U_{s} = ½ × 130 × 0.25

U_{s} = 16.25 J

Therefore, the elastic potential energy stored in a spring is **16.25 J**.

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**Related:**

- Thermal Energy Equation
- Potential Energy Formula
- Gravitational Potential Energy Formula
- Electric Potential Energy Formula
- Kinetic Energy Formula
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