A 2500-lb car can go from 0 to 88.0 ft/s (60 mph) in 8.00 s on a level road. The
ID: 1500058 • Letter: A
Question
A 2500-lb car can go from 0 to 88.0 ft/s (60 mph) in 8.00 s on a level road. The average drag force during the acceleration is 250 lbs. a) find the change in mechanical energy. answer: 3.035 x 105 ft lbs. b) determine the work done by the drag force. answer: -8.8 x 104 ft lbs c) Calculate the average power output of the engine. answer: 88.75 hp. d) Suppose this car takes on an extra 550-lb load and cruises up a 6% grade (measured using the old system where run is the inclined length) at 60 mph. Find the power required from the engine to go up this incline. answer: 68.8 hp.
Explanation / Answer
(a) change in mechanical energy= 1/2mv2-1/2mu2 w=1/2 x 2500/32.17 x 88x 88
w1 =3.009 x 10^5 ftlbs
(b) work done against drage force = F x S
S= ut + 1/2 at2
S = 0 + 1/2 x 11 x 64
S = 350 ft
W = - 250 x 352
W2 = - 8.8 x 10^4 ftlbs
(c) total work done = w1+w2
power = w1+w2/8 x 550.000037
= 388900/8 x 550.000037
= 88.38 hp
(d) I am not getting the meaning of 6% grade. If you can make it clear than I can try this out.
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