Answer: Calculate Average Power Output for 600 kg Top Fuel Drag Racer

AI Thread Summary
To calculate the average power output of a 600 kg top fuel drag racer completing a quarter-mile in 4.4 seconds, the initial calculations for acceleration, force, and work are correct. The average power output was calculated as 1432.747 kW based on work done over time. However, an alternative approach using the change in kinetic energy (ΔKE) suggests that work should be calculated as ΔKE = 0.5 * m * v^2, leading to a different power output. This method simplifies the process and may yield a more accurate result. The discussion emphasizes the importance of using the correct formula for work in power calculations.
lpcampbe
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A top fuel drag racer with a mass of 600.0 kg completes a quarter-mile (402 m) drag race in a time of 4.4 s starting from rest. The car's final speed is 115 m/s. What is the engine's average power output? Neglect friction and air resistance.

I know:
Power = Work/ change in time
W = Force * change in distance
F = mass * acceleration
Vf = Vi + at

so i solve for a
115 m/s = 0 m/s + a(4.4 s)
a = 26.1364 m/s^2

then i solve for F
F = 600 kg * 26.1364 m/s^2
F = 15681.8182 N

then i solve for W
W =15681.8182 N * 402 m
W = 6304090.909 J

then i solve for P
P = 6304090.909 J / 4.4 s
P = 1432747.934 W

I'm supposed to answer in kW
P = 1432.747 kW

I don't understand why this answer is wrong!
 
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lpcampbe said:
A top fuel drag racer with a mass of 600.0 kg completes a quarter-mile (402 m) drag race in a time of 4.4 s starting from rest. The car's final speed is 115 m/s. What is the engine's average power output? Neglect friction and air resistance.

I know:
Power = Work/ change in time
W = Force * change in distance
F = mass * acceleration
Vf = Vi + at

so i solve for a
115 m/s = 0 m/s + a(4.4 s)
a = 26.1364 m/s^2

then i solve for F
F = 600 kg * 26.1364 m/s^2
F = 15681.8182 N

then i solve for W
W =15681.8182 N * 402 m
W = 6304090.909 J

then i solve for P
P = 6304090.909 J / 4.4 s
P = 1432747.934 W

I'm supposed to answer in kW
P = 1432.747 kW

I don't understand why this answer is wrong!

Perhaps you want to consider that

Work = \Delta KineticEnergy = \frac{m*v^2}{2} - 0
 
Thanks, LowlyPion!

I also tried to find a using Vf^2 = Vi^2 + 2ax and i got this same answer, but the equation you gave me made my life so much easier. :)

Thanks again!
 
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