# Elastic Potential Energy Formula | Problems With Solutions

Elastic potential energy (Us) is half times the product of spring constant (k) and the square of displacement (Δx). Using the formula of elastic potential energy: Us = ½ × k Δx2, the value of elastic potential energy of spring can be calculated. (Δx is +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, Us = ?
Initial length of a spring, xi = 0.15 m
Finial length of a spring, xf = 0.30 m
Displacement of spring, Δx = xf – xi = 0.15 m
Spring constant, k = 150 N/m

Using the formula of elastic potential energy,
Us = ½ × k Δx2
Us = ½ × 150 × (0.15)2
Us = ½ × 150 × 0.0225
Us = 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, xi = 5 m
Finial length of a spring, xf = 2 m
Displacement of spring, Δx = xf – xi = -3 m
Elastic potential energy of a spring, Us = ?
Spring constant, k = 120 N/m

Using the formula of elastic potential energy,
Us = ½ × k Δx2
Us = ½ × 120 × (-3)2
Us = ½ × 120 × 9
Us = 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, xi = 0.25 m
Finial length of a spring, xf = 0.45 m
Displacement of spring, Δx = xf – xi = 0.2 m
Spring constant, k = 180 N/m
Elastic potential energy of a spring, Us = ?

Using the formula of elastic potential energy,
Us = ½ × k Δx2
Us = ½ × 180 × (0.2)2
Us = 90 × 0.04
Us = 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, xi = 0.85 m
Finial length of a spring, xf = 0.35 m
Displacement of spring, Δx = xf – xi = -0.5 m
Elastic potential energy of a spring, Us = ?
Spring constant, k = 130 N/m

Using the formula of elastic potential energy,
Us = ½ × k Δx2
Us = ½ × 130 × (-0.5)2
Us = ½ × 130 × 0.25
Us = 16.25 J

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

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