This preview shows pages 1–3. Sign up to view the full content.
Bell, Evan – Homework 5 – Due: Feb 18 2005, 11:00 pm – Inst: Furlan
1
This printout should have 32 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
be±ore answering.
The due time is Central
time.
001
(part 1 o± 1) 10 points
A cart loaded with bricks has a total mass
o± 21 kg and is pulled at constant speed by
a rope.
The rope is inclined at 27
.
2
◦
de
grees above the horizontal and the cart moves
17
.
3 m on a horizontal sur±ace. The coe²
cient o± kinetic ±riction between ground and
cart is 0
.
587.
The acceleration o± gravity is 9
.
8 m
/
s
2
.
How much work is done on the cart by the
rope?
Correct answer: 1605
.
56 J.
Explanation:
T
27
.
2
◦
Summing the components o± the ±orces
along the vertical direction provides
X
F
y
=
F
sin
θ
+
N 
mg
= 0
For the horizontal direction we have
X
F
x
=
F
cos
θ
+
μ
k
N
= 0
N
=
F
cos
θ
μ
k
,
so that
F
sin
θ
+
F
cos
θ
μ
k

mg
= 0
F μ
k
sin
θ
+
F
cos
θ
=
μ
k
mg
F
=
μ
k
mg
μ
k
sin
θ
+ cos
θ
.
Since
μ
k
sin
θ
+ cos
θ
= 0
.
587sin27
.
2
◦
+ cos27
.
2
◦
= 1
.
15773
,
the work done on the cart is given by
W
F
=
F d
cos
θ
=
μ
k
mg d
cos
θ
μ
k
sin
θ
+ cos
θ
=
0
.
587(21 kg)(9
.
8 m
/
s
2
)(17
.
3 m)cos27
.
2
◦
1
.
15773
= 1605
.
56 J
,
002
(part 1 o± 5) 10 points
A crate is pulled up a rough incline.
The
pulling ±orce is parallel to the incline. The
crate is pulled a distance o± 5
.
79 m.
The acceleration o± gravity is 9
.
8 m
/
s
2
.
8
91
kg
μ
=
0
292
142
N
1
48
m
/
s
23
.
4
◦
What is the magnitude o± the work is done
by the gravitational ±orce?
Correct answer: 200
.
787 J.
Explanation:
Let :
d
= 5
.
79 m
,
θ
= 23
.
4
◦
,
m
= 8
.
91 kg
,
g
= 9
.
8 m
/
s
2
,
μ
= 0
.
292
,
and
v
= 1
.
48 m
/
s
.
F
μ
N
N
mg
v
θ
The gravitational ±orce directed down the
plane is
F
grav
=

mg
sin
θ
, and the motion
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentBell, Evan – Homework 5 – Due: Feb 18 2005, 11:00 pm – Inst: Furlan
2
is directed up the plane, so the work by the
gravity is
W
grav
=

mg d
sin
θ
=

(8
.
91 kg)
g
(5
.
79 m)sin23
.
4
◦
=

200
.
787 J

W
grav

= 200
.
787 J
.
003
(part 2 of 5) 10 points
The work done by the gravitational ±eld is
1.
zero.
2.
positive.
3.
negative.
correct
Explanation:
see Part 1.
004
(part 3 of 5) 10 points
How much work is done by the 142 N force?
Correct answer: 822
.
18 J.
Explanation:
W
=
F d
= (142 N)(5
.
79 m)
= 822
.
18 J
,
where
d
is the distance along the incline.
005
(part 4 of 5) 10 points
What is the change in kinetic energy of the
crate?
Correct answer: 485
.
908 J.
Explanation:
W
f
=

f d
=

135
.
485 J
.
Δ
K
=
W
grav
+
W
app
+
W
f
= 485
.
908 J
.
006
(part 5 of 5) 10 points
What is the speed of the crate after it is pulled
5
.
79 m?
Correct answer: 10
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '07
 GALLIS,MICHAELRO
 Work

Click to edit the document details