Spring Stretch Calculation

Checking Comprehension

  • Topic: Elasticity and Hooke's Law

  • Overview of Hooke's Law: Hooke's Law describes the relationship between the force applied to a spring and the amount it stretches or compresses. It states that the force (F) exerted by a spring is directly proportional to the displacement (x) from its equilibrium position.

    • Mathematical Representation: The law can be represented mathematically as: F=kimesxF = k imes x where:

      • F: Force applied to the spring (in Newtons, N)

      • k: Spring constant (in Newtons per meter, N/m)

      • x: Displacement of the spring from its rest position (in meters, m)

  • Given Values in the Problem:

    • Spring constant, k = 65.5 N/m

    • Applied force, F = 15.3 N

  • Objective: Determine how far the spring will stretch (x).

    • Using Hooke's Law:

      1. Rearrange Hooke's Law to solve for displacement (x):
        x=Fkx = \frac{F}{k}

      2. Substitute the known values:
        x=15.3 N65.5 N/mx = \frac{15.3 \, N}{65.5 \, N/m}

      3. Calculate:
        x=15.365.5x = \frac{15.3}{65.5} = 0.2334 \, m (approximately)

  • Conclusion: The spring will stretch approximately 0.2334 meters (or 23.34 cm) when a force of 15.3 N is applied to it, based on the provided spring constant of 65.5 N/m.

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