(6%) Problem 17: A car engine produces a constant power of p = 350 hp. The car a
ID: 1874867 • Letter: #
Question
(6%) Problem 17: A car engine produces a constant power of p = 350 hp. The car and motor have a combined mass of m-2150 kg. The provided ansmar is not comeact and no specic feedback is aalable pjeose vie volsble hrts and nevew te relevant material cars speed, in meters per second? You may assume no energy is required to spin the tires. 50% Part (b) A top fuel dragster of mass m 1000.0 kg can accelerate and cover 305 m in 4.13 secs from a dead stop. How much horsepower is the engine producing at the wheels at the end of the race, assuming the acceleration is constant? Grade Summary P 162.78 Potentia % sin cos 7 8 9 HOME Attempts remaining: atan) acotan sinho % per attempt) detailed view Degrees Radians RACKSPACE Dil.| CLEAR | 4% Submit Hints: for a 0% deductian. Hints remaining: 0 Feedback: 1 for a 0% deduction You need to find the instantaneous power at the very last moment of the race. Recall the instantaneous power is the dot produce of the force and the velocity. What is the acceleration of this car? Be careful with the unit conversion to horsepower No meaningful feed back available for the current Submission. Submission History Answer Hints Feedback Totals 0% Note: Feedback lot 0% 4% -You need to find the instantaneous power at the very last moment of the race. Recall the instantaneous power is dot produce of the force and the velocity. What is the acceleration of this car? Be careful with the unit conversion to horsepower. theExplanation / Answer
(a)
350 Hp = 350*0.746 = 261.1 kw
Assuming constant power :
E = P*t = 261.1*10^3*4.13 = 1.08*10^6 joule = ½*mV^2 = 1075*V^2
V = (1.08*10^6/1075) = 31.7 m/sec
(b)
a = 2x/t^2 = (2*305)/4.13^2 = 35.76 m/sec^2
V max = a*t = 35.76*4.13 = 147.7 m/sec
Ek = 1/2mV^2 = 500*147.7^2 = 1.09*10^7 joule
P = Ek/t = 1.09*10^7/4.3 = 2.54 x 10^6 W = 3401.7 hp
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.