A 450 kg dragster accelerates from rest to a final speed of 110 m/s in 400 m (ab

Question

A 450 kg dragster accelerates from rest to a final speed of 110 m/s in 400 m (about a quarter of a mile) and encounters an average frictional force of 1200 N. What is its average power output in watts and horsepower if this takes 7.30 s How long would it take a 1.50 * 10^5 kg airplane with engines that produce a maximum of 100 MW of power to reach a speed of 200 m/s and an altitude of 12.0 km if air resistance were negligible If it actually takes 850 s, what is the power applied, in megawatts Given this power, what is the average force of air resistance if the airplane takes 1200 s

Explanation / Answer

Power P = net Work Done/time

P = KE - Ff /t

P = (0.5 mv^2 - 1200)/7.3

P = ((0.5 * 450 * 110*110) - 1200) /7.3

P = 372780.82Watts or 3.72 e 5 Watts

P = 372780.82/746 = 499.7 Hp or 500 Hp

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total energy = KE + PE

TE = 0.5 mv^2 + mgh

TE = 0.5 * 1.5 e 5 * 200*200 +(1.5 e5 * 9.81* 12)

TE = 3.017 e 9 J

Power P = Energy/time

time = E/P = 3.017 e9 /100 e 6

time t = 30.17 secs

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Power P = 3.017 e 9./850

P = 3.55 e 6 Watts or 3.56 MW

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let average force of air resistance is F N,

then work done against air resistance=force*distanc = F* vt

=F*1200 *200

then total energy supplied by the engine= 3.01 e 9 + F*240000

given that the engine is supplying power for 1200 seconds at 3.56 MW,

2.439*10^10+F*240000= 3.56*10^6 * 1200

F = 83.8 kN

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