Elastic Potential Energy Physics Example 4

Follow the full solution, then compare it with the other examples linked below.

Example 4

hard
A 0.2 kg0.2 \text{ kg} ball is launched vertically by a spring (k=500 N/mk = 500 \text{ N/m}) compressed by 0.1 m0.1 \text{ m}. How high does the ball rise above the release point? Use g=9.8 m/s2g = 9.8 \text{ m/s}^2.

Solution

  1. 1
    Elastic PE: PEspring=12(500)(0.01)=2.5 JPE_{\text{spring}} = \frac{1}{2}(500)(0.01) = 2.5 \text{ J}.
  2. 2
    At max height, all energy is gravitational PE: mgh=2.5mgh = 2.5.
  3. 3
    h=2.5mg=2.50.2×9.8=2.51.96≈1.28 mh = \frac{2.5}{mg} = \frac{2.5}{0.2 \times 9.8} = \frac{2.5}{1.96} \approx 1.28 \text{ m}

Answer

h≈1.28 mh \approx 1.28 \text{ m}
Energy conservation connects the spring's elastic PE directly to the ball's maximum gravitational PE. This avoids needing to track the ball's velocity at intermediate points.

About Elastic Potential Energy

Energy stored in an elastic object that has been stretched or compressed from its natural length.

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