johnson (mjj622) – HW 6 – Coker – (56625)
1
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printout
should
have
20
questions.
Multiplechoice questions may continue on
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before answering.
Homework on Chapter 7, which introduces
the concepts of work, kinetic energy, gravi
tational potential energy, conservation of en
ergy, and inclusion of nonconservative forces
such as friction in using energy concepts.
A
lot of stuff!
001
10.0 points
An object with mass
M
is attached to the end
of a string and is being lowered vertically at a
constant acceleration of
g
8
.
If it has been lowered a distance
ℓ
from rest,
how much work has been done by the tension
in the string?
1.
7
M g ℓ
8
2.

M g ℓ
4
3.

7
M g ℓ
8
correct
4.

M g ℓ
5.
M g ℓ
4
6.
M g ℓ
8
7.

M g ℓ
8
8.
M g ℓ
Explanation:
Using Newton’s Second Law,
F
net
=
T
+
M
g

ˆ
j
M a
=
T

ˆ
j
M g
T
=
ˆ
j
parenleftbigg
M g

M g
8
parenrightbigg
=
ˆ
j
parenleftbigg
7
M g
8
parenrightbigg
,
so the work done when Δ
x
=

ˆ
j
ℓ
is
W
=
F
·
Δ
x
=
parenleftbigg
7
M g
8
parenrightbigg
(

ℓ
) =

7
M g ℓ
8
.
002
(part 1 of 4) 10.0 points
A block of mass 1
.
12 kg is pushed 1
.
45 m along
a frictionless horizontal table by a constant
16 N force directed 27
◦
below the horizontal.
1
.
12 kg
μ
= 0
16 N
27
◦
Find the work done by the applied force.
Correct answer: 20
.
6713 J.
Explanation:
Consider the force diagram
F
θ
m g
n
f
k
Given :
m
= 1
.
12 kg
,
Δ
x
= 1
.
45 m
,
f
= 16 N
,
and
θ
= 27
◦
.
Only the horizontal component of the ap
plied force is used to move the block.
The total work due to the applied force is
W
x
=
f
x
Δ
x
=
f
cos
θ
Δ
x
= (16 N) (cos27
◦
) (1
.
45 m)
=
20
.
6713 J
.
003
(part 2 of 4) 10.0 points
Find the work done by the normal force ex
erted by the table.
Correct answer: 0 J.
Explanation:
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johnson (mjj622) – HW 6 – Coker – (56625)
2
Vertically, Δ
y
= 0
,
so
W
g
=
0 J
.
004
(part 3 of 4) 10.0 points
Find the work done by the force of gravity.
Correct answer: 0 J.
Explanation:
Vertically, Δ
y
= 0
,
so
W
g
=
0 J
.
005
(part 4 of 4) 10.0 points
Find the work done by the net force on the
block.
Correct answer: 20
.
6713 J.
Explanation:
The net force is
F
net
=
f
cos
θ
so the work done by
F
net
is
W
=
f
cos
θ
Δ
x
= (16 N) (cos 27
◦
) (1
.
45 m)
=
20
.
6713 J
.
006
10.0 points
An applied
force varies with position ac
cording to
F
=
k
1
x
n

k
2
, where
n
= 3,
k
1
= 6
.
8 N
/
m
3
, and
k
2
= 63 N.
How much work is done by this force on
an object that moves from
x
i
= 5
.
26 m to
x
f
= 18
.
7 m?
Correct answer: 205
.
733 kJ.
Explanation:
Basic Concepts:
W
=
integraldisplay
vector
F
·
dvectors
Solution:
The work done by a varying force
is
W
=
integraldisplay
x
2
x
1
vector
F
·
dvectors .
which adds up all the little
vector
F
·
vectors
parts along
the path, taking into account the changing
force. Here all the motion is in the ˆ
ı
direction
so
ds
=
dx
.
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 Fall '08
 Turner
 Energy, Force, Kinetic Energy, Potential Energy, Work, Coker

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