Spring Force Physics Example 3

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Example 3

medium
A spring with k=200 N/mk = 200 \text{ N/m} is compressed by 0.15 m0.15 \text{ m}. Find the spring force and the elastic potential energy stored.

Solution

  1. 1
    Apply Hooke's law for the restoring force: F=kx=200×0.15=30 NF = kx = 200 \times 0.15 = 30 \text{ N} (directed outward, opposing the compression).
  2. 2
    Calculate the elastic potential energy: PE=12kx2=12(200)(0.15)2=12(200)(0.0225)PE = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.15)^2 = \frac{1}{2}(200)(0.0225)
  3. 3
    PE=2.25 JPE = 2.25 \text{ J}

Answer

F=30 N,PE=2.25 JF = 30 \text{ N}, \quad PE = 2.25 \text{ J}
A compressed spring stores elastic potential energy proportional to the square of the displacement. The restoring force is linear in displacement (Hooke's law), but the energy grows quadratically.

About Spring Force

The restoring force exerted by a spring, proportional to how much it's stretched or compressed.

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