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Question (35)
Force by the road on a truck as it slows down.
A 1200-kg truck travels in a straight line and has a speed 25 m/s in the +x direction when the driver hits
the brakes. The truck slows down to 5 m/s in 10 s. Assume that the net force on the truck is constant.
(a) What is the force by the road and air on the truck as it slows down?
(b) What is the displacement of the truck during this time interval?
Solution
(a) Begin by drawing a picture, showing the initial momentum, final momentum, and the change in momentum of the truck.
Figure 1: A truck slows down with a constant net force.
Figure 2: The change in momentum for the truck.
Apply the momentum principle. Here are the steps:
(1) define the system:
System
= truck
(2) apply the momentum principle to the system. The truck’s initial and final momentum and the
time interval must have been measured in order to calculate the net force.
∆~
p = F~net ∆t
∆~
p
F~net =
∆t
m∆~v
F~net =
∆t
m(~
vf − ~vi )
F~net =
∆t
Substitute the given quantities
F~net
m(~vf − ~vi )
∆t
(1200 kg)(h5, 0, 0i − h25, 0, 0i) m/s
=
10 s
(1200 kg)(h−20, 0, 0i) m/s
=
10 s
= h−2400, 0, 0iN
=
Check your answer. Note that the net force points to the left, in the –x direction. This is consistent with
the direction of the change in momentum from our picture.
To calculate an individual force acting on the system, you must apply the definition of net force as the vector
sum of all forces acting on the system. Here are the steps:
(1) List the objects in the surroundings that exert forces on the system. This list includes any
objects that makes contact with the system and any objects that exerts “force at a distance” such as a
gravitational force or an electrostatic force or a magnetic force, for example. Here are the objects in the
surroundings that exert forces on the system.
objects that interact with the system:
truck
air (which is significant for a moving truck)
road
Earth (via gravitational force)
(2) Sketch a force diagram showing the forces acting on the object with approximate magnitudes and directions if they are known. Air resistance is certainly not negligible; however, we do not
have enough information to distinguish the force by air on the truck and the force by the road on the truck.
Therefore, we will combine the force by air and the force by the road into a single force that represents the
sum of the force by air and the force by the road. However, we’ll refer to it below as the force by the road
on the truck, even though it includes the effect of air.
Figure 3: The force diagram showing the forces on the truck.
Note that the force by the road on the truck must be in the direction shown so that when adding the vectors
tail to head, the resultant (or net force) is in the direction given by the momentum principle. Verify that
the forces in the force diagram sum to give the correct net force.
Figure 4: The sum of the forces on the truck give the correct net force.
(3) Apply the definition of net force and solve for the unknown force. Sometimes you have to do
this in component form and sometimes you can do it in vector form. In this case, we’ll use vector form.
F~net
F~by
road on truck
= F~grav by Earth on truck + F~by road
= F~net − F~grav by Earth on truck
on truck
= h−2400, 0, 0iN − (1200 kg)(< 0, −9.8, 0 > N/kg)
= h−2400, 0, 0iN− < 0, −11760, 0 > N
= h−2400, 11760, 0iN
The force by the road on the truck points upward and to the left, exactly as shown in our free-body diagram.
The vertical component of the force by the road on the truck balances the gravitational force on the truck
so that the net force on the truck in the y-direction is zero. The horizontal component of the force by the
road on the truck causes the truck to slow down. Note that the brakes are internal to the system. Therefore,
the brakes do not exert a force on the truck. It is the road and air that are responsible for slowing down the
truck (when the brakes are pressed or if it is coasting) or speed up the truck (when the gas pedal is pressed
enough to cause the truck to speed up).
(b) To find the displacement of the truck, use the definition of average velocity and solve for the displacement
of the truck.
~vavg
∆~r
∆~r
∆t
= ~vavg ∆t
=
The average velocity can be found by using the arithmetic mean of the initial and final velocity of the truck.
~vavg
~vi + ~vf
2
< 5, 0, 0 > m/s+ < 25, 0, 0 > m/s
=
2
< 30, 0, 0 > m/s
=
2
= < 15, 0, 0 > m/s
=
Substitute the average velocity and solve for the displacement.
∆~r
= ~vavg ∆t
= (< 15, 0, 0 > m/s)(10 s)
= < 150, 0, 0 > m
The truck travels 150 m as it slows down. That’s longer than a football field. As you can imagine this is a
fairly small acceleration. The truck is perhaps coasting to a stop rather than applying its brakes.