Thevalingam, Donald – Homework 14 – Due: Dec 21 2006, midnight – Inst: Eslami
1
This printout should have 18 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
Please take note oF the due date oF the as
signment. It is due on Thursday, Decembber
21 at midnight.
001
(part 1 oF 1) 10 points
An elevator starts From rest with a constant
upward acceleration and moves 1 m in the frst
1
.
7 s. A passenger in the elevator is holding a
5
.
5 kg bundle at the end oF a vertical cord.
The acceleration oF gravity is 9
.
8 m
/
s
2
.
What is the tension in the cord as the ele
vator accelerates?
Correct answer: 57
.
7062 N.
Explanation:
T
mg
a
elevator
g
Let
h
be the distance traveled and
a
the
acceleration oF the elevator. With the initial
velocity being zero, we simpliFy the Following
expression and solve For acceleration oF the
elevator:
h
=
v
0
t
+
1
2
at
2
=
1
2
at
2
=
⇒
a
=
2
h
t
2
.
The equation describing the Forces acting on
the bundle is
F
net
=
ma
=
T

mg
T
=
m
(
g
+
a
)
=
m
µ
g
+
2
h
t
2
¶
= (5
.
5 kg)
•
9
.
8 m
/
s
2
+
2(1 m)
(1
.
7 s)
2
‚
= 57
.
7062 N
.
keywords:
002
(part 1 oF 3) 10 points
A dragster and driver together have mass
881
.
2 kg
.
The dragster, starting From rest,
attains a speed oF 26
.
8 m
/
s in 0
.
54 s
.
±ind the average acceleration oF the drag
ster during this time interval.
Correct answer: 49
.
6296 m
/
s
2
.
Explanation:
The average acceleration during time
t
is
a
=
Δ
v
t
=
v

0
t
=
v
t
=
(26
.
8 m
/
s)
(0
.
54 s)
=
49
.
6296 m
/
s
2
.
003
(part 2 oF 3) 10 points
What is the size oF the average Force on the
dragster during this time interval?
Correct answer: 43733
.
6 N.
Explanation:
The average Force on the dragster is
f
d
=
m
d
a
= (881
.
2 kg)(49
.
6296 m
/
s
2
)
=
43733
.
6 N
.
004
(part 3 oF 3) 10 points
Assume:
The driver has a mass oF 73
.
2 kg
.
What horizontal Force does the seat exert
on the driver?
Correct answer: 3632
.
89 N.
Explanation:
The Force on the driver is
F
=
ma
= (73
.
2 kg)(49
.
6296 m
/
s
2
)
=
3632
.
89 N
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThevalingam, Donald – Homework 14 – Due: Dec 21 2006, midnight – Inst: Eslami
2
keywords:
005
(part 1 of 2) 10 points
The tension in a string from which a 7
.
4 kg
object is suspended in an elevator is equal to
40 N.
The acceleration of gravity is 9
This is the end of the preview.
Sign up
to
access the rest of the document.