Transportation
Energy Use in Cars 3: Rolling Friction
Problem Set Solutions
Problem 1: The coefficient of rolling resistance for a shipping truck is typically around 0.01. Because
train wheels are much harder and deform much less during travel, the coefficient of rolling resistance
for a train is around 0.001. Assuming both the truck and train have diesel engines with 37% efficiency,
how much extra diesel will be burned by the truck to transport 1 tonne load for 1 km?
Solution 1:
Approach: Calculate the force of rolling resistance for both the truck and the train. Calculate the energy
consumption and fuel cost for each kilometre travelled. Calculate the amount of diesel per kilometre.
Find the difference between the two values.
What we know:
1 tonne = 1 x 103 kg
coefficient of rolling resistance of truck RR(truck) = 0.01
coefficient of rolling resistance of train RR(train) = 0.001
efficiency = 37%
Energy content of diesel = 32 MJ/L
Note: to simplify this calculation the mass of the truck and train have not been included.
Calculate force of rolling resistance (FRR):
FRRtruck  μ RR mg
 (0.01)(1  10 3 kg)(9.8 m/s 2 )
 100 N
FRRtrain  μ RR mg
 (0.001)(1  10 3 kg)(9.8 m/s 2 )
 10 N
Calculate the energy of driving one kilometre:
Wtruck  FRR d
 (100 N)(1000 m)
 100,000N  m
 100 kJ
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Wtrain  FRR d
 (10 N)(1000 m)
 10,000N  m
 10 kJ
Using the efficiency, calculate the fuel energy input required for each kilometre:
Work Output
Efficiency
100 kJ

37%
 270 kJ
Fuel Energy Input (truck) 
Work Output
Efficiency
10 kJ

37%
 27 kJ
Fuel Energy Input (train) 
Using the energy content of diesel we can figure out how much diesel is used for a truck and a train to
travel one kilometre.
# of litres (truck) 
# of joules
Energy per litre
270 kJ
32x10 3 kJ/L
 0.0084 L

# of litres (train) 
# of joules
Energy perlitre
27 kJ
32x10 3 kJ/L
 0.00084 L/km

Difference in fuel between the truck and the train is 0.0084 L – 0.00084 L = 0.00756 L for every km of
travel.
Since the coefficient of rolling resistance is larger by a factor of 10 for the truck, the energy consumption
and volume of fuel used is also higher by a factor of 10. Furthermore in favour of the train is the fact
that the train cars ride in the slip stream of the locomotive which yields another advantage if we were to
take air resistance into account.
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Problem 2: Lightening the Load
One of the recommendations for saving energy and fuel is to remove all unnecessary things from the
trunk of your car and reduce its total weight. How much gas can you save over a driving distance of 100
km if you reduce the weight of your car by 50 kg?
The car’s owner’s manual tells you that the empty weight of your car is 1300 kg and the car is 2 m wide
and 1.4 m high. You know that modern gasoline engines are around 25 % efficient at converting
chemical energy into mechanical energy and gasoline has 32 MJ/L of chemical energy. You look up the
drag coefficient on your iPhone (CD = 0.28) and you recall from your physics class that the coefficient of
rolling resistance for a car is ~0.02.
Problem 2 Solution: Lightening the Load
What we know:
d = 100 km
wremoved = 50 kg
wcar = 1300 kg
Acar = 2 m x 1.4 m = 2.8 m2
η = 25%
energy content of gasoline = 32 MJ/L
CD = 0.28
μRR 0.02
We know that work is defined to be the forces opposing the motion of the car, rolling friction and
air drag, multiplied by distance.
W  Fd
 (Fr  Fd )d
 Fr d  Fd d
1
  RR mgd  C D Av 2 d
2
You could do this calculation by assuming an average speed and calculating the fuel consumption before
and after emptying the trunk. However, since the mass doesn’t affect air drag, the work done by the
drag force will be the same before and after emptying the trunk. The only thing that changes is the
force due to rolling friction:
Fr   RR mg
There is a 3.5% different in weight, from 1430 kg to 1380 kg (including the weight of the driver),
meaning the work done by the rolling friction is also changed by 3.5%, as is the fuel consumption.
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Before:
Wr 
1

 RR mgd
1
(0.02)(143 0 kg)(9.8 m/s 2 )(100,000 m)
0.25
 112 MJ

L : 112 MJ
1L
 3.5 L
32 MJ
After:
Wr 
1

 RR mgd
1
(0.02)(138 0 kg)(9.8 m/s 2 )(100,000 m)
0.25
 108 MJ

L : 108 MJ
1L
 3.38 L
32 MJ
The 3.5% difference in the weight corresponds to a fuel saving of just 0.12 L when driving 100 km.
Note: here we assume that the coefficient of rolling friction doesn’t change. In reality, tires will be a
little flatter at larger weight, leading also to a change in the coefficient of rolling friction.
Mathew (Sandy) Martinuk, Georg Rieger and Brittany Tymos 2010/08/03
Physics and Astronomy Outreach Program at the University of British Columbia