Conservation of Energy Physics Example 4

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Example 4

hard
A skier (60 kg60 \text{ kg}) starts from rest at the top of a 25 m25 \text{ m} hill and reaches the bottom at 18 m/s18 \text{ m/s}. How much energy was lost to friction? Use g=10 m/s2g = 10 \text{ m/s}^2.

Solution

  1. 1
    Initial energy (all PE): Ei=mgh=60×10×25=15,000 JE_i = mgh = 60 \times 10 \times 25 = 15{,}000 \text{ J}.
  2. 2
    Final energy (all KE): Ef=12mv2=12(60)(324)=9720 JE_f = \frac{1}{2}mv^2 = \frac{1}{2}(60)(324) = 9720 \text{ J}.
  3. 3
    Energy lost to friction: Efriction=Ei−Ef=15,000−9720=5280 JE_{\text{friction}} = E_i - E_f = 15{,}000 - 9720 = 5280 \text{ J}.

Answer

Efriction=5280 JE_{\text{friction}} = 5280 \text{ J}
When friction is present, mechanical energy is not fully conserved — some is converted to thermal energy. The difference between initial and final mechanical energy tells us exactly how much was lost.

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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