Power: Practice Problem & Solution
A car of mass \(1000\text{ kg}\) moving at \(20\text{ m/s}\) is brought to rest in \(5\text{ s}\). The average power dissipated due to braking is
Solution Explained:
To solve this problem, we apply the core principles of Power. Understanding the underlying formula is key to arriving at the correct answer below:
The initial kinetic energy of the car is \(K_i = \frac{1}{2}mv^2 = \frac{1}{2}(1000)(20)^2 = 2 \times 10^5\text{ J}\) and final kinetic energy is zero. Average power dissipated is the work done divided by time: \(P = \frac{\Delta K}{t} = \frac{2 \times 10^5\text{ J}}{5\text{ s}} = 40\text{ kW}\).
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