Version 072 – TEST02 – tsoi – (92940)
1
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printout
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have
16
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before answering.
001
10.0points
Normally the rate at which you expend energy
during a brisk walk is 3
.
1 Calories per minute.
(A Calorie is the common unit of food energy,
equal to 4184
.
1 J.)
How long do you have to walk to produce
the same amount of energy as a 40 W light
bulb that is lit for 6
.
9 hours?
1. 1.7208
2. 2.18685
3. 0.834327
4. 1.99891
5. 7.35868
6. 1.29523
7. 3.54726
8. 6.6681
9. 1.69816
10. 1.27672
Correct answer: 1
.
27672 h.
Explanation:
Let :
P
= 3
.
1 Cal
/
min
and
t
= 6
.
9 h
.
The energy expended by the light in 6
.
9 h
is
E
=
P t
= (40 W) (6
.
9 h)
×
3600 s
h
= 9
.
936
×
10
5
J
.
The rate at which energy is being used while
walking is
P
′
= (3
.
1 Cal
/
min) (4184
.
1 J
/
Cal)
= 12970
.
7 J
/
min
,
so the time need to produce the same amount
of energy is
t
′
=
E
P
′
=
9
.
936
×
10
5
J
12970
.
7 J
/
min
·
1 h
60 min
=
1
.
27672 h
.
002
10.0points
To test the performance of its tires, a car
travels along a perfectly flat (no banking) cir
cular track of radius 389 m. The car increases
its speed at uniform rate of
a
t
≡
d

v

dt
= 4
.
64 m
/
s
2
until the tires start to skid.
If the tires start to skid when the car reaches
a speed of 32 m
/
s, what is the coefficient of
static friction between the tires and the road?
The acceleration of gravity is 9
.
8 m
/
s
2
.
1. 0.355893
2. 0.63605
3. 0.582742
4. 0.323145
5. 0.544358
6. 0.559861
7. 0.509574
8. 0.620436
9. 0.406725
10. 0.759253
Correct answer: 0
.
544358.
Explanation:
Let :
R
= 389 m
,
v
= 32 m
/
s
,
and
g
= 9
.
8 m
/
s
2
.
We need the car’s acceleration just before
the skidding occurs. The car travels in a circle
rather than along a straight line, so it has the
centripetal acceleration
a
c
=
v
2
R
.
In addition, the car increases its speed, so
there is a tangential acceleration
a
t
≡
d

v

dt
= 4
.
64 m
/
s
2
.
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Version 072 – TEST02 – tsoi – (92940)
2
The net acceleration of the car is the vector
sum
vector
a
=
vector
a
c
+
vector
a
t
;
since the tangential and the centripetal accel
erations have perpendicular orientations, the
magnitude of the net acceleration is

a

=
radicalBig
a
2
c
+
a
2
t
=
radicalbigg
v
4
R
2
+
a
2
t
=
radicalBigg
(32 m
/
s)
4
(389 m)
2
+ (4
.
64 m
/
s
2
)
2
= 5
.
33471 m
/
s
2
.
The force providing this acceleration is the
static friction force
vector
F
s
between the tires and
the track – the only horizontal force available
on the flat track.
The static friction force
may have any horizontal direction, and any
magnitude up to the maximum of
F
max
s
=
μ
s
N
=
μ
s
(
M g
)
.
Consequently, the net horizontal acceleration
of the car is limited by
M

a
 ≤
μ
s
M g

a
 ≤
μ
s
g .
When the car starts to exceed this accelera
tion limit, the tires skid, so the static friction
coefficient is
μ
s
=

a

max
g
=
5
.
33471 m
/
s
2
9
.
8 m
/
s
2
=
0
.
544358
.
003
10.0points
The system of point masses shown in the
figure
is rotating at an
angular speed
of
ω
= 1
.
41 rev
/
s.
The masses (all equal) are
connected by light, flexible spokes that can be
lengthened or shortened.
At the beginning,
each spoke is
r
= 0
.
879 m long.
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