# hw18_s - homework 18 YOO HEE Due Mar 1 2008 4:00 am F = ma...

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homework 18 – YOO, HEE – Due: Mar 1 2008, 4:00 am 1 Question 1, chap 8, sect 5. part 1 of 1 10 points A hot rod of mass 1600 kg, starting from rest reaches a speed of 150 m / s in only 18 . 4 s. What is the average output power? 1. 0 . 978261 MW correct 2. 0 . 000354442 MW 3. 1 . 95652 MW 4. 0 . 0130435 MW 5. 0 . 0531664 MW Explanation: The average output power is equal to the kinetic energy acquired divided by the total time, so P = 1 2 mv 2 t = 1 2 (1600 kg) (150 m / s) 2 18 . 4 s · mW 10 6 W = 0 . 978261 MW Question 2, chap 8, sect 5. part 1 of 1 10 points A car with a mass of 1 . 61 × 10 3 kg starts from rest and accelerates to a speed of 11.5 m/s in 10.8 s. Assume that the force of resistance remains constant at 450.7 N during this time. What is the average power developed by the car’s engine? Correct answer: 12449 W (tolerance ± 1 %). Explanation: Basic Concepts: v f = a Δ t and Δ x = 1 2 a t ) 2 since v i = 0 m/s. P = W Δ t = F Δ x Δ t F = ma + F r Given: m = 1 . 61 × 10 3 kg v f = 11 . 5 m / s Δ t = 10 . 8 s F r = 450 . 7 N Solution: a = v f Δ t so P = F Δ x Δ t = F · 1 2 a t ) 2 Δ t = Fa Δ t 2 = ( ma + F r ) v f 2 = p m v f Δ t + F r P · v f 2 = b (1610 kg)(11 . 5 m / s) 10 . 8 s + 450 . 7 N B · 11 . 5 m / s 2 = 12449 W Question 3, chap 8, sect 5. part 1 of 1

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hw18_s - homework 18 YOO HEE Due Mar 1 2008 4:00 am F = ma...

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