How am I supposed to do part b of this problem? Suppose a car travels 108 km at

Question

How am I supposed to do part b of this problem?

Suppose a car travels 108 km at a speed of 20.0 m/s, and uses 1.90 gallons of gasoline. Only 30% of the gasoline goes into work done against friction. (The energy content of gasoline is 1.3 ? 108 J per gallon.)

(a) What is the force of friction?
686.1 Correct.

(b) If friction is directly proportional to speed, how many gallons will be used to drive 108 km at a speed of 28.0 m/s?
  
2.65 Incorrect: Your answer is incorrect.
gallons

Please show work. Thank you so much!

Explanation / Answer

displacement x = 108 km = 108000 m

f2 / f1 = v2/v1

f2 = (f1*v2)/v1

f2 = (686.1*28)/20 = 960.54 N

work done by frictional force Wf = f*x = 960.54*108000 = 103738320J

number of gallons used = Wf/(0.3*1.3*10^8) = 2.66 gallons

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