Energy Physics Example 2

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Example 2

medium
A 1000 kg1000 \text{ kg} car traveling at 20 m/s20 \text{ m/s} brakes to a stop. How much energy is dissipated as heat?

Solution

  1. 1
    Initial kinetic energy: KE=12mv2=12(1000)(400)=200,000 JKE = \frac{1}{2}mv^2 = \frac{1}{2}(1000)(400) = 200{,}000 \text{ J}
  2. 2
    Final kinetic energy: KEf=0KE_f = 0 (car at rest).
  3. 3
    Energy dissipated: ΔE=200,000−0=200,000 J=200 kJ\Delta E = 200{,}000 - 0 = 200{,}000 \text{ J} = 200 \text{ kJ}

Answer

ΔE=200 kJ\Delta E = 200 \text{ kJ}
When brakes stop a car, kinetic energy is converted to thermal energy through friction. Energy is conserved overall but is transformed from useful kinetic energy to less useful heat.

About Energy

The capacity to do work or cause change in a physical system, measured in joules (J).

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