Conservation of Energy Physics Example 3

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Example 3

medium
A pendulum of length 2 m2 \text{ m} is released from a height 0.5 m0.5 \text{ m} above its lowest point. What is its speed at the lowest point? Use g=10 m/s2g = 10 \text{ m/s}^2.

Solution

  1. 1
    At the top: KE=0KE = 0, PE=mgh=m(10)(0.5)=5mPE = mgh = m(10)(0.5) = 5m.
  2. 2
    At the bottom: PE=0PE = 0, KE=12mv2KE = \frac{1}{2}mv^2.
  3. 3
    Conservation: 5m=12mv2  ⟹  v=10≈3.16 m/s5m = \frac{1}{2}mv^2 \implies v = \sqrt{10} \approx 3.16 \text{ m/s}.

Answer

v≈3.16 m/sv \approx 3.16 \text{ m/s}
The mass cancels out in energy conservation problems involving only gravity. The speed at the bottom depends only on the height dropped.

About Conservation of Energy

A fundamental law of physics stating that the total energy of an isolated system remains constant over time — energy can be transferred between objects.

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