Question

12) There are several types of drag on a car other than air resistance. Effects having to do with the squeezing of the tires (rolling resistance) and frictional forces in the drivetrain (the system that transfers energy from the engine to the rotation of the wheels) also must be taken into account. Engineers use the following equation to model the total force due to these different effects

Fdrag=A+Bv+Cv2Fdrag=A+Bv+Cv2

For a Camry, these coefficients are estimated to be A=117.130A=117.130 N, B=1.800 N s/mB=1.800 N s/m, and C=0.368 N s2/m2C=0.368 N s2/m2.

Suppose that the driver steadily accelerates the car from 0 km/hr to 100 km/hr over a 3.3 s. What is the magnitude of the work done by the drag forces?

Hint: What is the instantaneous power of the drag? What is the relationship between that and the total work done by drag forces? Which kinematic model can you use to relate velocity, time, and acceleration?

How much energy is lost to drag when the car travels 20 km at 29 m/s?

Hint: Figure out the power due to the drag. How do you use this to figure out total work when velocity is constant? Which kinematic model is most useful for relating time, velocity, and displacement in this situation?

The car drives a distance 120 m up a section of road whose grade is 25% (the grade is the vertical rise of the road as a percent of the horizontal run of the road). The car's speed is 29 m/s. The car has a mass of 1543.60 kg. What is the minimum amount of power needed for the car to drive up this section of road, assuming perfect efficiency?

Hint: Grade of the road is height divided by horizontal distance. Model the road as a slanted slope and the rise and run as the base and height of a right triangle whose hypotenuse is the distance driven by the car. Why might the height the car rises during this section of road be important from the perspective of energy? What is the power associated with gravitational work?

Answer 1

part a:

drag force=A+B*v+C*v^2

instantaneous power=integral of F*dv

=integral of (A+B*v+C*v^2)*dv

=A*v+0.5*B*v^2+(C*v^3/3)

work done=integration of power*dt

final velocity=100 km/hr=100*1000/3600

=27.778 m/s

acceleration =(final velocity-initial velocity)/total time=8.4175 m/s^2

as v=8.4175*t

work done =integration of (A*8.4175*t+0.5*B*8.4175^2*t^2+(C*(8.4175*t)^3/3))*dt

=492.97*t^2+21.256*t^3+18.29*t^4

using the limits from t=0 to 3.3 seconds,

work done=8301.4 J

part b:

at v=29 m/s, drag force=A+B*v+C*v^2

=478.82 N

work done against drag force=force*distance

=9576.4 J

part c:

mass =m=1543.6 kg

angle with horizontal=arctan(0.25)=14.036 degrees

force on the car=mass*g*sin(theta)

so power required=force*speed

=106398.225 W