fullwood (tf5349) – Homework 6 Dynamics – Catala – (49033)
1
This printout should have 21 questions.
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beFore answering.
001
(part 1 oF 2) 10.0 points
A block oF mass 3
.
26 kg lies on a Frictionless
horizontal surFace.
The block is connected
by a cord passing over a pulley to another
block oF mass 7
.
47 kg which hangs in the air,
as shown on the Following picture. Assume
the cord to be light (massless and weightless)
and unstretchable and the pulley to have no
Friction and no rotational inertia.
The acceleration oF gravity is 9
.
8 m
/
s
2
.
3
.
26 kg
7
.
47 kg
Calculate the acceleration oF the frst block.
Correct answer: 6
.
82255 m
/
s
2
.
Explanation:
Given :
m
1
= 3
.
26 kg
,
and
m
2
= 7
.
47 kg
.
m
1
m
2
a
T
N
m
1
g
a
T
m
2
g
Since the cord is unstretchable, the frst
block accelerates to the right at exactly the
same rate
a
as the second (hanging) block ac
celerates downward. Also, the cord’s tension
pulls the frst block to the right with exactly
the same tension
T
as it pulls the second block
upward.
The only horizontal Force acting on the frst
block is the cord’s tension
T
, hence by New
ton’s Second Law
m
1
a
=
F
net
→
1
=
T .
The second block Feels two vertical Forces:
The cord’s tension
T
(upward) and the block’s
own weight
W
2
=
m
2
g
(downward). Conse
quently,
m
2
a
=
F
net
↓
2
=
m
2
g
−
T .
Adding the two equations together, we arrive
at
(
m
1
+
m
2
)
a
=
m
2
g ,
and hence
a
=
m
2
m
1
+
m
2
g
=
7
.
47 kg
3
.
26 kg + 7
.
47 kg
(9
.
8 m
/
s
2
)
= 6
.
82255 m
/
s
2
.
002
(part 2 oF 2) 10.0 points
Calculate the tension in the cord.
Correct answer: 22
.
2415 N.
Explanation:
T
=
m
1
a
= (3
.
26 kg) (6
.
82255 m
/
s
2
)
= 22
.
2415 N
.
003
(part 1 oF 2) 10.0 points
A child holds a sled on a Frictionless, snow
covered hill, inclined at an angle oF 27
◦
.
79 N
F
27
◦
IF the sled weighs 79 N, fnd the Force ex
erted on the rope by the child.
Correct answer: 35
.
8653 N.
Explanation:
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View Full Documentfullwood (tf5349) – Homework 6 Dynamics – Catala – (49033)
2
Given :
W
= 79 N
and
θ
= 27
◦
.
Consider the free body diagram for the
block
mg
sin
θ
N
=
cos
F
W
Basic Concepts:
If we “tilt” our world,
and consider the forces parallel to the hill,
F
net
=
s
F
up
−
s
F
down
= 0
then the forces perpendicular to the hill,
F
net
=
s
F
out
−
s
F
in
= 0
Solution:
Consider the free body diagram
for the sled: The weight of the sled has compo
nents
W
sin
θ
acting down the hill and
W
cos
θ
acting straight into the hill.
The system is in equilibrium, so for forces
parallel to the hill,
F
net
=
T
−W
sin
θ
= 0
=
⇒
T
=
W
sin
θ
= (79 N) sin27
◦
= 35
.
8653 N
004
(part 2 of 2) 10.0 points
What force is exerted on the sled by the hill?
Correct answer: 70
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 Spring '10
 PRO.KIROSHIMA
 Force, Mass, Work, Correct Answer, Fnet, Newton’s Second Law, Dynamics – Catala

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