Conservation of Energy Physics Example 2

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Example 2

hard
A roller coaster car (500 kg500 \text{ kg}) starts from rest at 30 m30 \text{ m} high and descends to 10 m10 \text{ m}. What is its speed at 10 m10 \text{ m}? Use g=10 m/s2g = 10 \text{ m/s}^2.

Solution

  1. 1
    Energy at top: E1=mgh1=500×10×30=150,000 JE_1 = mgh_1 = 500 \times 10 \times 30 = 150{,}000 \text{ J}.
  2. 2
    At height 10 m10 \text{ m}: E2=12mv2+mgh2E_2 = \frac{1}{2}mv^2 + mgh_2.
  3. 3
    Conservation: 150,000=12(500)v2+500(10)(10)150{,}000 = \frac{1}{2}(500)v^2 + 500(10)(10).
  4. 4
    150,000=250v2+50,000  ⟹  250v2=100,000  ⟹  v=20 m/s150{,}000 = 250v^2 + 50{,}000 \implies 250v^2 = 100{,}000 \implies v = 20 \text{ m/s}

Answer

v=20 m/sv = 20 \text{ m/s}
Conservation of energy allows us to find speeds at any height without knowing the path details. Only the height difference matters for gravitational PE changes.

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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