A typical compact car experiences a total drag force at 55 m i/h of about 350 N. If this car gets 35 miles per gallon of gasoline at this speed, and a liter of gasoline (1 gal = 3.8 L) releases about 3.2 X 107J when burned, what is the car’s efficiency

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valetta valetta · Answer 1 · 2019-08-06T14:06:51+02:00

Answer:

Efficiency = 16.2124 %

Explanation:

Given:

velocity, v = 55 mi/hr = 24.5872 m/s

Drag force = 350 N

Work done by friction = Force x displacement

now for 1 second,

we get

Power (P) as

P = 350 × 24.5872 = 8605.52 W

Now,

The consumption of gasoline by car for travelling 35 miles = 3.8 L

Energy released by burning of 3.8 L of gasoline = 3.2 × 10⁷ J

Time taken to travel 35 mile = 35/55 = 0.6363 hour = 0.6363 × 3600 seconds = 2290.90 seconds

Thus,

The power used = (3.2 x 10⁷ × 3.8)/2290.90 = 53079.57 W

or

Therefore,

The efficiency = [output power/ input power] × 100%

The efficiency = [8605.52 W /53079.57] x 100%

Efficiency = 16.2124 %