Elastic potential energy formula

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Elastic potential energy formula
Elastic potential energy formula | Image: Learnool

The elastic potential energy formula, expressed as Us = ½ × k Δx2, establishes a relationship between the elastic potential energy (Us) of a spring, the spring constant (k), and the square of the displacement (Δx). By utilizing this formula, one can calculate the amount of elastic potential energy stored in a spring, taking into account the spring constant and the square of the displacement. It’s worth noting that the displacement (Δx) is positive when the spring is stretched and negative when the spring is compressed.

Practice problems

Problem #1

Calculate the elastic potential energy of a spring when it is stretched from a length of 0.15 m to 0.30 m. Use a value of the 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
  • Therefore, the displacement of spring, Δx = xf – xi = 0.15 m
  • Spring constant, k = 150 N/m

Applying the formula:

  • 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 initially measuring 5 m in length is compressed to a final length of 2 m. Determine the elastic potential energy of the spring. Use a value of the 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
  • Therefore, the displacement of spring, Δx = xf – xi = -3 m
  • Elastic potential energy of a spring, Us = ?
  • Spring constant, k = 120 N/m

Applying the formula:

  • 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

Find the elastic potential energy stored in a spring that is stretched from its rest length of 0.25 m to 0.45 m. The spring has a spring constant of k = 180 N/m.

Solution

Given data:

  • Elastic potential energy of a spring, Us = ?
  • Initial length of a spring, xi = 0.25 m
  • Finial length of a spring, xf = 0.45 m
  • Therefore, the displacement of spring, Δx = xf – xi = 0.2 m
  • Spring constant, k = 180 N/m

Applying the formula:

  • 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. Determine the elastic potential energy stored in the spring, when it is compressed to a length of 0.35 m. Use a spring constant value of k = 130 N/m.

Solution

Given data:

  • Elastic potential energy of a spring, Us = ?
  • Initial length of a spring, xi = 0.85 m
  • Finial length of a spring, xf = 0.35 m
  • Therefore, the displacement of spring, Δx = xf – xi = -0.5 m
  • Spring constant, k = 130 N/m

Applying the formula:

  • 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.

Related

External links

Deep

Learnool.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.

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