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Physics 221
Fall 2008
Homework #5 Solutions
Ch. 5:3,4,
Ch. 6
Due Tues, Sept 30, 2008
5.1 A car of mass
M
is moving in a straight line at an initial speed of
v
i
on a horizontal concrete
highway.
For rubber on dry concrete, the coefficient of static friction is 1.0 and the coefficient of
kinetic friction is 0.8.
For rubber on wet concrete, the coefficient of static friction is 0.30 and the
coefficient of kinetic friction is 0.25.
(a)
Draw a free body diagram for the forces on the car once the brakes on the car are applied.
Here,
f
is the frictional force of the road on the tires,
n
is the normal force of the road on the
tires, and
w
is the weight of the car.
The initial velocity of the car is to the right.
(b) Derive an expression for the (constant, negative) acceleration of the car after the brakes are
applied, in terms of the coefficient of friction
μ
of the car tires with the road.
Because there is no acceleration of the car normal to the road, the magnitude of the normal
force is just the weight
Mg
of the car.
Then
a
x
=
f
x
M
=
±
μ
n
M
=
±
(
Mg
)
M
=
±
g
(c)
From the answer in part (b), derive an expression for the distance that the car takes to come to
a stop once the brakes are applied.
This is a constant acceleration problem with
v
x
= 0 and
x
0
= 0.
Then using Eq. (2.13) one gets
x
=
±
v
0
x
2
2
a
x
=
±
v
0
x
2
2(
±
g
)
=
v
0
x
2
2
g
.
In the following assume that
v
i
= 67 mph (30 m/s).
(d)
Calculate the minimum stopping distance in meters for the car on dry
concrete, if the car is not
skidding (e.g., using antilock brakes).
x
=
v
0
x
2
2
s
g
=
(30 m/s)
2
2(1.0)(9.80 m/s
2
)
=
46 m
x
y
n
v
w
f
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(e)
Calculate the stopping distance in meters for the car on dry
concrete, if the tires are skidding
(e.g., if the brakes are locked).
x
=
v
0
x
2
2
μ
k
g
=
(30 m/s)
2
2(0.8)(9.80 m/s
2
)
=
57 m
(f)
Calculate the minimum stopping distance in meters for the car on wet
concrete, if the car is not
skidding (e.g., using antilock brakes).
x
=
v
0
x
2
2
s
g
=
(30 m/s)
2
2(0.30)(9.80 m/s
2
)
=
150 m
(g) Calculate the stopping distance in meters for the car on wet
concrete, if the tires are skidding
(e.g., if the brakes are locked).
x
=
v
0
x
2
2
k
g
=
(30 m/s)
2
2(0.25)(9.80 m/s
2
)
=
180 m
5.2 A 1.02 kg block of brass is sitting still on a steel inclined plane.
The inclined plane is at a small
angle
±
with respect to the horizontal.
The angle
±
is slowly increased until the block just starts to
slide down the inclined plane, and then
±
is held constant after that.
The coefficient of static
friction of brass on steel is 0.51, and the coefficient of kinetic friction of brass on steel is 0.44.
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 Fall '06
 Johnson
 Mass, Work

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