Thermal Energy Physics Example 4

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Example 4

hard
A 1200 kg1200 \text{ kg} car traveling at 30 m/s30 \text{ m/s} brakes to a stop. If all kinetic energy is converted to thermal energy in the brakes (4 kg4 \text{ kg} of steel, c=500 J/(kg\cdotp°C)c = 500 \text{ J/(kg·°C)}), by how much do the brakes heat up?

Solution

  1. 1
    Kinetic energy: KE=12(1200)(900)=540,000 JKE = \frac{1}{2}(1200)(900) = 540{,}000 \text{ J}.
  2. 2
    All KE becomes thermal energy in the brakes: Q=mcΔTQ = mc\Delta T.
  3. 3
    ΔT=Qmc=540,0004×500=540,0002000=270°C\Delta T = \frac{Q}{mc} = \frac{540{,}000}{4 \times 500} = \frac{540{,}000}{2000} = 270°\text{C}

Answer

ΔT=270°C\Delta T = 270°\text{C}
Braking converts kinetic energy to thermal energy through friction. This is why brakes can get extremely hot after heavy use, and why brake cooling is critical in motorsport and heavy vehicles.

About Thermal Energy

The total kinetic energy of all particles (atoms and molecules) in an object due to their random motion.

Learn more about Thermal Energy →

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