ouseph (jao888) – HW03 – tsoi – (92940)
1
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printout
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have
24
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0points
An elevator is being lifted up an elevator shaft
at a constant speed by a steel cable as shown
in the figure below.
All frictional effects are
negligible.
steel
cable
Elevator going up
at constant speed
In this situation, forces on the elevator are
such that
1.
the upward force by the cable is greater
than the downward force of gravity.
2.
the upward force by the cable is smaller
than the downward force of gravity.
3.
the upward force by the cable is greater
than the sum of the downward force of gravity
and a downward force due to the air.
4.
the upward force by the cable is equal to
the downward force of gravity.
correct
5.
None of these.
(The elevator goes up
because the cable is being shortened, not be
cause an upward force is exerted on the eleva
tor by the cable.)
Explanation:
Since the elevator is being lifted at a con
stant speed, the net force on it is zero, there
fore, the upward force by the cable is equal to
the downward force of gravity.
002
10.0points
Is it correct to say that no force acts on a
body at rest?
1.
All are wrong.
2.
No net force acts on a body at rest; when
the net force is zero, the body is in static
equilibrium.
correct
3.
No force acts on a body at rest; if at least
one force acted on it the body would move.
4.
No force acts on a body at rest; all forces
cancel each other.
5.
No net force acts on a body at rest; no
force acts on the body at all.
Explanation:
There may be any number of forces that
act to produce a zero net force. When the net
force is zero, the body is in static equilibrium.
003
10.0points
An elevator starts from rest with a constant
upward acceleration and moves 1 m in the first
1
.
6 s. A passenger in the elevator is holding a
7
.
4 kg bundle at the end of a vertical cord.
What is the tension in the cord as the ele
vator accelerates? The acceleration of gravity
is 9
.
8 m
/
s
2
.
Correct answer: 78
.
3013 N.
Explanation:
h
= 1 m
,
t
= 1
.
6 s
,
m
= 7
.
4 kg
,
and
g
= 9
.
8 m
/
s
2
.
T
mg
a
elevator
g
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ouseph (jao888) – HW03 – tsoi – (92940)
2
Let
h
be the distance traveled and
a
the
acceleration of the elevator. Since the initial
velocity is zero,
h
=
v
0
t
+
1
2
a t
2
=
1
2
a t
2
a
=
2
h
t
2
.
The equation describing the forces acting on
the bundle is
F
net
=
m a
=
T
−
m g
T
=
m
(
g
+
a
) =
m
parenleftbigg
g
+
2
h
t
2
parenrightbigg
= (7
.
4 kg)
bracketleftbigg
9
.
8 m
/
s
2
+
2 (1 m)
(1
.
6 s)
2
bracketrightbigg
=
78
.
3013 N
.
004(part1of3)10.0points
You are driving a car down a straight road at
a constant 60 miles per hour.
Consider the following forces:
I) air drag pushing back on the car;
II) gravity pulling down on the car;
III) the ground pushing up on the car;
IV) friction pushing the wheels (and the car)
forward; and
V) friction pushing the wheels (and the car)
back.
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 Summer '08
 Staff
 Physics, Force, kg

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