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Chapter 6: Problem 75

A top fuel drag racer with a mass of \(500.0 \mathrm{kg}\) completes a quarter- mile \((402 \mathrm{m})\) drag race in a time of \(4.2 \mathrm{s}\) starting from rest. The car's final speed is \(125 \mathrm{m} / \mathrm{s}\). What is the engine's average power output? Ignore friction and air resistance.

Short Answer

Expert verified
Answer: The average power output of the drag racer's engine is approximately 930,059.5 W.

Step by step solution

01

Determine the initial and final kinetic energy of the drag racer

To find the initial and final kinetic energies, we can use the formula for kinetic energy: \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass of the object and \(v\) its velocity. Given that the drag racer starts at rest, its initial velocity is \(0\mathrm{m}/\mathrm{s}\). For the initial kinetic energy: \(KE_i = \frac{1}{2} (500.0\mathrm{kg})(0\mathrm{m}/\mathrm{s})^2 = 0\mathrm{J}\) For the final kinetic energy, we are given that its final velocity is \(125\mathrm{m}/\mathrm{s}\): \(KE_f = \frac{1}{2}(500.0\mathrm{kg})(125\mathrm{m}/\mathrm{s})^2 = 3,906,250\mathrm{J}\)
02

Calculate the work done on the drag racer

We can use the work-energy principle to find the total work done on the drag racer. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy: \(W = KE_f - KE_i = 3,906,250\mathrm{J} - 0\mathrm{J} = 3,906,250\mathrm{J}\)
03

Calculate the average power output

Now that we have calculated the total work done on the drag racer, we can find the average power output by dividing the work done by the time taken: \(P_{avg} = \frac{W}{t} = \frac{3,906,250\mathrm{J}}{4.2\mathrm{s}} = 930,059.5\mathrm{W}\) So, the engine's average power output is approximately \(930,059.5\mathrm{W}\).

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