Spring Force Physics Example 4

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Example 4

medium
A spring (k=300 N/mk = 300 \text{ N/m}) is compressed by 0.15 m0.15 \text{ m}. What force does it exert, and how much elastic potential energy is stored?

Solution

  1. 1
    Force: F=kx=300×0.15=45 NF = kx = 300 \times 0.15 = 45 \text{ N} (pushing outward).
  2. 2
    Elastic PE: PE=12kx2=12(300)(0.15)2=12(300)(0.0225)=3.375 JPE = \frac{1}{2}kx^2 = \frac{1}{2}(300)(0.15)^2 = \frac{1}{2}(300)(0.0225) = 3.375 \text{ J}

Answer

F=45 N,PE=3.375 JF = 45 \text{ N}, \quad PE = 3.375 \text{ J}
A compressed spring exerts an outward restoring force following Hooke's law. The elastic potential energy stored is proportional to the square of the compression, so doubling the compression quadruples the stored energy.

About Spring Force

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

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