Physics Homework 4
pdf
keyboard_arrow_up
School
The University of Oklahoma *
*We aren’t endorsed by this school
Course
2414
Subject
Communications
Date
Feb 20, 2024
Type
Pages
3
Uploaded by EarlPuppy847
theExpertTA.com
|
Student: Jocelyn.D.Pranabdev-1@ou.edu
My Account
Log Out
Class Management
|
Help
Problem 1: Given Newton’s First Law of Motion, what do we reasonably expect an object to do given the following scenarios? Part (a)
An object sits at rest with no unbalanced forces acting upon it. What do we expect this object to do?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
The object will remain at rest.
The object will remain at rest.
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
12:24 PM
The object will remain at rest.
Part (b)
An object is traveling with a constant velocity with no unbalanced forces acting upon it. What do we expect this object to do?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
The object will remain at the same speed,
traveling in the same direction.
The object will remain at the same speed,
traveling in the same direction.
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
12:25 PM
The object will remain at the same speed, traveling in the same direction.
Part (c)
An object sits at rest with an unbalanced force acting on it. What do we expect this object to do?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
The object will begin to move with an
increasing velocity.
The object will begin to move with an
increasing velocity.
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
12:25 PM
The object will begin to move with an increasing velocity.
Problem 2: A boxer's fist and glove have a mass of m
= 0.82
kg. The boxer's fist can obtain a speed of v
= 6.75
m/s in a time of t
=
0.19
s. Part (a)
Find the magnitude of the average acceleration a
ave
, in meters per square second, of the boxer's fist.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
a
ave
= a
ave
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:21 PM
a
ave
= Part (b)
How much force did the boxer apply to his fist/glove, in newtons? Grade = 100%
Correct Answer
Student Final Submission
Feedback
F
b
= F
b
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:22 PM
F
b
= Problem 3: Consider a bowling ball of mass M
attached to two ropes. The bowling
ball is in equilibrium, with one rope tied to the ceiling and the second rope pulling
horizontally. Refer to the figure on the right. How is the tension T
2
related to the weight of the bowling ball?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
T
2
is greater than the bowling ball's weight.
T
2
is greater than the bowling ball's weight.
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints Feedback
1
Feb 16, 2023
2:22 PM
T
2
is greater than T
1
.
Note: Feedback provided here but not accessed during assignment.
You are absolutely right, T
2
is greater than T
1
, but this question asks you to
compare T
2
and force of gravity.
2
Feb 16, 2023
2:22 PM
T
2
is greater than the bowling
ball's weight.
Problem 4: In the figure, the net external force on the 24 kg mower is known to be 51
N.
Randomized Variables
f
= 25
N
v
= 1.1
m/s
Part (a)
If the force of friction opposing the motion is 25
N, what force F (in newtons) is the person exerting on the mower?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
F
= F
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:22 PM
F
= - 25
F
= 2
Feb 16, 2023
2:22 PM
F
= 3
Feb 16, 2023
2:23 PM
F
= Note: Feedback provided here but not accessed during assignment.
Check the free body diagram: force of friction opposes the motion of the mower.
4
Feb 16, 2023
2:23 PM
F
= - 26
F
= 5
Feb 16, 2023
3:16 PM
F
= Part (b)
Suppose the mower is moving at 1.1
m/s when the force F is removed. How far will the mower go before stopping in m?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
x
= x
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
3:18 PM
x
= Problem 5: A box rests on a rough horizontal surface. You apply a force, F
, on the box
at an angle, θ
, as shown in the figure and it slides at a constant velocity. If necessary,
use Fs
and Fk
for the forces of static and kinetic friction. Please use the interactive area below to draw the Free Body Diagram for this block, assuming it is in equilibrium.
Grade = 0%
Give Up was accessed and a deduction for using the Give Up may have been applied.
Correct Answer
Student Final Submission
Feedback
There is no "velocity" force.
Velocity does not belong in a
Free Body Diagram.
You are missing at least one
force. It looks like there is at least
one force that is expected in
the drawing, but isn't in the
right direction. Grade Summary
Deduction for Final Submission
100
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
2
%
Student Grade = 100 - 100
- 2 = 0%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints Feedback
1
Feb 16, 2023
2:29 PM
Note: Feedback provided here but not
accessed during assignment.
The object is moving. Please think about what
frictional forces should be considered here.
You are missing at least one force. It looks like there is at least one force that is
expected in the drawing, but isn't in the right
direction. 2
Feb 16, 2023
2:29 PM
Note: Feedback provided here but not
accessed during assignment.
The object is moving. Please think about what
frictional forces should be considered here.
You are missing at least one force. It looks like there is at least one force that is
expected in the drawing, but isn't in the right
direction. 3
Feb 16, 2023
3:22 PM
Note: Feedback provided here but not
accessed during assignment.
The object is moving. Please think about what
frictional forces should be considered here.
You are missing at least one force and you have
some extra forces that do not belong in the
diagram. 4
Feb 16, 2023
8:57 PM
Note: Feedback provided here but not
accessed during assignment.
The object is moving. Please think about what
frictional forces should be considered here.
You are missing at least one force. It looks like there is at least one force that is
expected in the drawing, but isn't in the right
direction. 5
Feb 16, 2023
8:57 PM
Note: Feedback provided here but not
accessed during assignment.
The object is moving. Please think about what
frictional forces should be considered here.
You are missing at least one force. 6
Feb 16, 2023
8:57 PM
Note: Feedback provided here but not
accessed during assignment.
The object is moving. Please think about what
frictional forces should be considered here.
You are missing at least one force. 7
Feb 16, 2023
9:52 PM
Note: Feedback provided here but not
accessed during assignment.
You are missing at least one force. 8
Feb 16, 2023
9:53 PM
Note: Feedback provided here but not
accessed during assignment.
There is no "acceleration" force. Acceleration
does not belong in a Free Body Diagram.
You are missing at least one force. It looks like there is at least one force that is
expected in the drawing, but isn't in the right
direction. 9
Feb 16, 2023
9:53 PM
There is no "velocity" force. Velocity does not
belong in a Free Body Diagram.
You are missing at least one force. It looks like there is at least one force that is
expected in the drawing, but isn't in the right
direction. Problem 6: A block with a mass of m
= 15
kg rests on a frictionless surface and is
subject to two forces acting on it. The first force is directed in the negative x
-direction
with a magnitude of F
1
= 8.5
N. The second has a magnitude of F
2
= 18
N and acts on
the body at an angle θ
= 17
° measured from horizontal, as shown. Part (a)
Please select the correct free body diagram from the choices below.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints Feedback
1
Feb 16, 2023
2:29 PM
Note: Feedback provided here but not accessed during assignment.
You are missing at least one force. What else can cause a force on the body that you might
be forgetting? 2
Feb 16, 2023
2:29 PM
Note: Feedback provided here but not accessed during assignment.
You are missing at least one force. Have you considered all forces that would be imparted
on the body by interaction with other objects?
3
Feb 16, 2023
2:29 PM
Note: Feedback provided here but not accessed during assignment.
You are missing at least one force, perhaps more. When drawing an FBD you must
consider ALL forces acting on an object, not just the externally applied forces. Are there
reaction forces that are caused by interaction of the body with other surfaces or objects?
Are there other forces that are caused by the body itself. 4
Feb 16, 2023
2:29 PM
Note: Feedback provided here but not accessed during assignment.
You are missing at least one force. Look closely at the provided image and the forces that
are specified or that may arise as a result of applied forces.
5
Feb 16, 2023
2:29 PM
Part (b)
Find the block's acceleration in the x
-direction, a
x
, in meters per second squared.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
a
x
= a
x
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:32 PM
a
x
= Note: Feedback provided here but not accessed during assignment.
You may have forgotten to divide by the mass.
2
Feb 16, 2023
2:32 PM
a
x
= - 8.71
a
x
= 3
Feb 16, 2023
2:34 PM
a
x
= 4
Feb 16, 2023
2:34 PM
a
x
= Note: Feedback provided here but not accessed during assignment.
You may have forgotten to divide by the mass.
5
Feb 16, 2023
2:34 PM
a
x
= Note: Feedback provided here but not accessed during assignment.
You may have forgotten to divide by the mass.
6
Feb 16, 2023
2:34 PM
a
x
= Note: Feedback provided here but not accessed during assignment.
You may have forgotten to divide by the mass.
7
Feb 16, 2023
2:34 PM
a
x
= 9.0
a
x
= 8
Feb 16, 2023
2:34 PM
a
x
= 9
Feb 16, 2023
2:34 PM
a
x
= 10 Feb 16, 2023
2:34 PM
a
x
= Note: Feedback provided here but not accessed during assignment.
You may have forgotten to divide by the mass.
11 Feb 16, 2023
2:34 PM
a
x
= 12 Feb 16, 2023
2:34 PM
a
x
= Note: Feedback provided here but not accessed during assignment.
You may have forgotten to divide by the mass.
13 Feb 16, 2023
2:34 PM
a
x
= 8.50
a
x
= Note: Feedback provided here but not accessed during assignment.
You may have forgotten to divide by the mass.
14 Feb 16, 2023
2:34 PM
a
x
= Note: Feedback provided here but not accessed during assignment.
You may have forgotten to divide by the mass.
15 Feb 16, 2023
2:35 PM
a
x
= Note: Feedback provided here but not accessed during assignment.
You may have forgotten to divide by the mass.
16 Feb 16, 2023
2:35 PM
a
x
= Note: Feedback provided here but not accessed during assignment.
You may have forgotten to divide by the mass.
17 Feb 16, 2023
2:35 PM
a
x
= Note: Feedback provided here but not accessed during assignment.
You may have forgotten to divide by the mass.
18 Feb 16, 2023
7:19 PM
a
x
= 19 Feb 16, 2023
7:35 PM
a
x
= Problem 7: A block having a mass of m
= 16
kg is suspended via two cables as shown
in the figure. The angles shown in the figure are as follows: α
= 13
° and β
= 35
°. We
will label the tension in Cable 1 as T
1
and the tension in Cable 2 as T
2
. Part (a)
From the images below, choose the correct free body diagram.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints Feedback
1
Feb 16, 2023
2:35 PM
Note: Feedback provided here but not accessed during assignment.
The two tensions pictured here are not along the directions of the ropes. Please examine
the geometry of the problem closely.
2
Feb 16, 2023
2:35 PM
Note: Feedback provided here but not accessed during assignment.
While it is acceptable to break forces down into thier components, the y component of
tension two has been neglected here. 3
Feb 16, 2023
2:35 PM
Note: Feedback provided here but not accessed during assignment.
The forces here are rotated unnecessarily. The tensions would act along the direction of
the ropes shown in the main figure.
4
Feb 16, 2023
2:35 PM
Note: Feedback provided here but not accessed during assignment.
The forces here are mirrored unnecessarily. The tensions would act along the direction of
the ropes shown in the main figure. Please examine the geometry closely.
5
Feb 16, 2023
2:35 PM
Part (b)
Write an expression for the sum of forces in the x direction in terms of T
1
, T
2
, m
, g
, α
, and β
. Use the specified coordinate system. Grade = 100%
Correct Answer
Student Final Submission
Feedback
Σ
F
x
= T
2
cos(
β
) - T
1
sin(
α
)
Σ
F
x
= T
2
cos(
β
) - T
1
sin(
α
)
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:37 PM
Σ
F
x
= T
2
cos(
β
) - T
1
sin(
α
)
Part (c)
Write an expression for the sum of forces in the y
direction in terms of T
1
, T
2
, m
, g
, α
, and β
. Use the specified coordinate system. Grade = 100%
Correct Answer
Student Final Submission
Feedback
Σ
F
y
= T
2
sin(
β
) + T
1
cos(
α
) - m g
Σ
F
y
= T
1
cos(
α
) + T
2
sin(
β
) - m g
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:38 PM
Σ
F
y
= T
1
cos(
α
) + T
2
sin(
β
) - m g
Part (d)
Solve for the numeric value of T
1
, in newtons.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
T
1
= T
1
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
7:49 PM
T
1
= 204.87
T
1
= 2
Feb 16, 2023
7:50 PM
T
1
= - 204.87
T
1
= 3
Feb 16, 2023
8:29 PM
T
1
= 4
Feb 16, 2023
9:22 PM
T
1
= 5
Feb 16, 2023
9:22 PM
T
1
= 6
Feb 16, 2023
9:22 PM
T
1
= Part (e)
Solve for the numeric value of T
2
, in newtons.
Grade = 0%
Give Up was accessed and a deduction for using the Give Up may have been applied.
Correct Answer
Student Final Submission
Feedback
T
2
= T
2
= 103.34
T
2
= Grade Summary
Deduction for Final Submission
100
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
8
%
Student Grade = 100 - 100
- 8 = 0%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
7:50 PM
T
2
=
204.87
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
2
Feb 16, 2023
7:50 PM
T
2
=
- 204.87
T
2
= 3
Feb 16, 2023
7:50 PM
T
2
=
205.87
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
4
Feb 16, 2023
8:30 PM
T
2
= 5
Feb 16, 2023
9:22 PM
T
2
= 6
Feb 16, 2023
9:22 PM
T
2
= 7
Feb 16, 2023
9:22 PM
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
8
Feb 16, 2023
9:22 PM
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
9
Feb 16, 2023
9:22 PM
T
2
=
- 203.9
T
2
= 10 Feb 16, 2023
9:22 PM
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
11 Feb 16, 2023
9:23 PM
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
12 Feb 16, 2023
9:23 PM
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
13 Feb 16, 2023
9:23 PM
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
14 Feb 16, 2023
9:23 PM
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
15 Feb 16, 2023
9:23 PM
T
2
= 16 Feb 16, 2023
9:23 PM
T
2
= 17 Feb 16, 2023
9:23 PM
T
2
= 18 Feb 16, 2023
9:23 PM
T
2
= 19 Feb 16, 2023
9:23 PM
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
20 Feb 16, 2023
9:23 PM
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
21 Feb 16, 2023
9:23 PM
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
22 Feb 16, 2023
9:23 PM
T
= Note: Feedback provided here but not accessed
35.53
35.53
35.53
29.13
29.13
29.13
76.00
76.00
-
25.00
25.00
26.00
-
26.00
76.00
0.5808
0.5800
0.5800
θ
Fn
Fg
F
Fk
θ
Fg
Fn
Fk
v
Fs
Fn
θ
Fs
Fn
θ
Fs
Fn
Fn
Fg
Fs
F
θ
Fs
F
θ
F
Fk
Fg
Fn
Fk
θ
Fg
Fn
Fk
a
θ
Fg
Fn
Fk
v
0.5809
0.5800
8.710
-
8.710
0.000
8.730
8.800
8.600
9.000
8.200
8.300
8.500
8.400
8.550
8.500
8.590
8.580
8.750
8.790
15.06
0.5800
138.7
142.0
204.9
-
204.9
683.6
144.4
144.0
142.0
38.08
103.3
204.9
-
204.9
205.9
683.6
144.0
157.0
203.0
203.9
-
203.9
206.0
207.0
208.0
209.0
210.0
215.0
214.0
213.0
212.0
211.0
202.0
201.0
22 Feb 16, 2023
9:23 PM
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
23 Feb 16, 2023
9:23 PM
T
2
= 24 Feb 16, 2023
9:25 PM
T
2
= 25 Feb 16, 2023
9:25 PM
T
2
= 26 Feb 16, 2023
9:25 PM
T
2
= 27 Feb 16, 2023
9:25 PM
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have mixed up one or more of the angles
associated with the trigonometric terms. Carefully
review which angles go with which trigonometric terms.
28 Feb 16, 2023
9:28 PM
T
2
=
163.91
T
2
= 29 Feb 16, 2023
9:38 PM
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
30 Feb 16, 2023
9:38 PM
T
2
=
- 202.2
T
2
= 31 Feb 16, 2023
9:39 PM
T
2
=
202.22
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
32 Feb 16, 2023
9:39 PM
T
2
=
202.27
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
33 Feb 16, 2023
9:39 PM
T
2
=
204.13
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
34 Feb 16, 2023
9:39 PM
T
2
=
204.19
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
35 Feb 16, 2023
9:39 PM
T
2
=
204.89
T
2
= You may have confused tan and cotan in your formulas.
Check your trigonometry.
36 Feb 16, 2023
9:55 PM
T
2
=
191.75
T
2
= 37 Feb 16, 2023
9:55 PM
T
2
=
202.77
T
2
= You may have confused tan and cotan in your formulas.
Check your trigonometry.
38 Feb 16, 2023
9:55 PM
T
2
= 39 Feb 16, 2023
9:55 PM
T
2
= 40 Feb 16, 2023
9:55 PM
T
2
= 41 Feb 16, 2023
9:55 PM
T
2
= 42 Feb 16, 2023
9:56 PM
T
2
= 43 Feb 16, 2023
9:56 PM
T
2
= 44 Feb 16, 2023
9:56 PM
T
2
= 45 Feb 16, 2023
9:57 PM
T
2
= 46 Feb 16, 2023
9:57 PM
T
2
= 47 Feb 16, 2023
9:57 PM
T
2
= 48 Feb 16, 2023
9:57 PM
T
2
= 49 Feb 16, 2023
9:57 PM
T
2
= 50 Feb 16, 2023
9:57 PM
T
2
=
182.034
T
2
= 51 Feb 16, 2023
9:57 PM
T
2
=
209.89
T
2
= You may have confused tan and cotan in your formulas.
Check your trigonometry.
52 Feb 16, 2023
9:57 PM
T
2
= 53 Feb 16, 2023
9:57 PM
T
2
= -Use your expressions from parts (b) and (c).
Since you know the answer for T
1
, you can
plug that into either equation to find T
2
.
54 Feb 16, 2023
9:58 PM
T
2
=
204.45
T
2
= Note: Feedback provided here but not accessed
during assignment.
You may have confused tan and cotan in your formulas.
Check your trigonometry.
55 Feb 16, 2023
9:58 PM
T
2
= 56 Feb 16, 2023
9:58 PM
T
2
=
103.34
T
2
= Problem 8: A gymnast with a mass of 45
-kg jumps on a trampoline. What force does a trampoline have to apply to accelerate her straight up at 7.1
m/s
2
in Newtons?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
F
t
= F
t
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:41 PM
F
t
= Note: Feedback provided here but not accessed during assignment.
Use the proper number of parentheses in your calculations
2
Feb 16, 2023
2:42 PM
F
t
= Problem 9: Attached to the rear-view mirror of a car is a small crystal of mass 50 g on a
string. When the car is stopped at a light, the crystal hangs vertically. When the light turns
green, the driver accelerates and notices the crystal makes an angle of θ
= 11
degrees with
respect to the vertical. Part (a)
Please select the correct free body diagram, using an inertial coordinate system fixed to the road, given F
g
is the force due to gravity, F
T
is
the tension in the string, F
a
is the centrifugal acceleration and F
N
is the normal force.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:42 PM
Part (b)
Input an expression for the magnitude of the car's acceleration in terms of the variables provided - the acceleration due to gravity and θ
.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
a
= g tan(
θ
)
a
= g tan(
θ
)
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:44 PM
a
= g tan(
θ
)
Part (c)
What is the car's acceleration in m/s
2
?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
a
= a
= 1.90
a
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:44 PM
a
= 1.90
a
= Part (d)
When the car is no longer accelerating, what is the new angle, θ
2
in degrees?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
θ
2
= θ
2
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:45 PM
θ
2
= Problem 10: An elevator filled with passengers has a mass of 1950
kg. Part (a)
The elevator accelerates upward (in the positive direction) from rest at a rate of 1.9
m/s
2
for 2.3
s. Calculate the tension in the cable
supporting the elevator in newtons.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
T
= T
= 22815
T
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:53 PM
T
= 22815
T
= Part (b)
The elevator continues upward at constant velocity for 8.1
s. What is the tension in the cable, in newtons, during this time?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
T
= T
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:54 PM
T
= Part (c)
The elevator experiences a negative acceleration at a rate of 0.55
m/s
2
for 2.8
s. What is the tension in the cable, in newtons, during this
period of negative acceleration?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
T
= T
= 18037.5
T
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:54 PM
T
= 18037.5
T
= Part (d)
How far, in meters, has the elevator moved above its original starting point?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
y
= y
= 50.50
y
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints Feedback
1
Feb 16, 2023
2:55 PM
y
= 2
Feb 16, 2023
2:56 PM
y
=
- 2.185
y
= 3
Feb 16, 2023
2:58 PM
y
=
5.0255
y
= Note: Feedback provided here but not accessed during assignment.
The answer provided was not correct. We have recognized the following,
- Your answer appears to be off by a factor of 10
n
, where n is an integer value. Ensure you have
represented the number in the correct units.
4
Feb 16, 2023
2:58 PM
y
=
4.2555
y
= 5
Feb 16, 2023
2:58 PM
y
= 6
Feb 16, 2023
3:00 PM
y
= 7
Feb 16, 2023
3:00 PM
y
= 8
Feb 16, 2023
3:01 PM
y
= 9
Feb 16, 2023
3:06 PM
y
= 10 Feb 16, 2023
3:06 PM
y
=
- 8.45
y
= 11 Feb 16, 2023
3:06 PM
y
= 12 Feb 16, 2023
3:06 PM
y
= Note: Feedback provided here but not accessed during assignment.
The answer provided was not correct. We have recognized the following,
- Your answer appears to be off by a factor of 10
n
, where n is an integer value. Ensure you have
represented the number in the correct units.
13 Feb 16, 2023
3:06 PM
y
= - 5.1
y
= Note: Feedback provided here but not accessed during assignment.
The answer provided was not correct. We have recognized the following,
- Your answer appears to be off by a factor of -10
n
, where n is an integer value. Ensure you have
represented the number in the correct units. Your answer contains a sign error.
14 Feb 16, 2023
3:11 PM
y
= 55.0
y
= Note: Feedback provided here but not accessed during assignment.
One of the acclerations is negative - you have not correctly included this.
15 Feb 16, 2023
3:11 PM
y
= 16 Feb 16, 2023
3:11 PM
y
= Note: Feedback provided here but not accessed during assignment.
One of the acclerations is negative - you have not correctly included this.
17 Feb 16, 2023
3:14 PM
y
=
50.50
y
= Problem 11: The figure shows Superhero and Trusty Sidekick hanging motionless
from a rope. Superhero’s mass is 88
kg, while Trusty Sidekick’s is 53
kg, and the mass
of the rope is negligible.
Randomized Variables
m
1
= 88
kg
m
2
= 53
kg
Part (a)
Find the tension in the rope above Superhero in Newtons.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
T
= T
= 1381.8
T
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:48 PM
T
= - 2763.6
T
= Note: Feedback provided here but not accessed during assignment.
The answer provided was not correct. We have recognized the following,
- Your answer appears to be off by a factor of -2. Your answer contains a sign error.
2
Feb 16, 2023
2:48 PM
T
= 2763.6
T
= Note: Feedback provided here but not accessed during assignment.
The answer provided was not correct. We have recognized the following,
- Your answer appears to be off by a factor of 2. 3
Feb 16, 2023
2:48 PM
T
= - 1381.8
T
= Note: Feedback provided here but not accessed during assignment.
The answer provided was not correct. We have recognized the following,
- Your answer appears to be off by a factor of -1. Your answer contains a sign error.
4
Feb 16, 2023
2:48 PM
T
= 1381.8
T
= Part (b)
Find the tension in the rope between Superhero and Trusty Sidekick in Newtons.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
T'
= T'
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
2:50 PM
T'
= Problem 12: A car and a dump truck are involved in an accident and crash into each other. Assuming the only force acting is the force of collision, which one experiences the most force during the crash - the car or the dump truck?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
They experience the same size force.
They experience the same size force.
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints Feedback
1
Feb 16, 2023
2:46 PM
The dump truck.
Note: Feedback provided here but not accessed during assignment.
The car pushing on the truck and the truck pushing on the car form an
action-reaction pair.
2
Feb 16, 2023
2:46 PM
It depends on which one is
moving faster.
Note: Feedback provided here but not accessed during assignment.
The car pushing on the truck and the truck pushing on the car form an
action-reaction pair.
3
Feb 16, 2023
2:46 PM
They experience the same size
force.
Problem 13: A car and a dump truck are involved in an accident and crash into each other. Which one will have the greatest magnitude of acceleration during the crash? (Assume the dump truck has a mass that is 4 times larger than the car,
and the only forces in the problem are those between the two vehicles.)
Grade = 100%
Correct Answer
Student Final Submission
Feedback
The car.
The car.
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints Feedback
1
Feb 16, 2023
2:46 PM
It depends on
the angle of
their collision.
Note: Feedback provided here but not accessed during assignment.
Regardless of the angle of their collisions, Newton's Third Law still holds true (i.e. the
forces acting on each are equal and opposite). By Newton's Second Law the force on an
object is related to the mass times acceleration. 2
Feb 16, 2023
2:46 PM
The dump truck.
Note: Feedback provided here but not accessed during assignment.
According to Newton's Third Law This the magnitude of the forces exerted on the car
and the dumptruck are equal. By Newton's Second Law the force on an object is related
to the mass times acceleration. 3
Feb 16, 2023
2:46 PM
It depends on
which one is
moving faster.
Note: Feedback provided here but not accessed during assignment.
Regardless of the angle of their respective speeds, Newton's Third Law still holds true
(i.e. the forces acting on each are equal and opposite). By Newton's Second Law the
force on an object is related to the mass times acceleration. 4
Feb 16, 2023
2:46 PM
The car.
Problem 14: A teenager of mass pushes backward against the ground with his foot as he rides his skateboard
towards the right. This exerts a horizontal force of magnitude . The skateboard has mass . Part (a)
Write an expression for the magnitude of the horizontal component of the force,
, that the ground exerts on the teenager's foot.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
= F
foot
= F
foot
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
7:57 PM
= ( m
1
+ m
2
) a
2
Feb 16, 2023
7:57 PM
= a ( m
1
+ m
2
)
3
Feb 16, 2023
7:58 PM
= F
N
15.2
4
Feb 16, 2023
7:58 PM
= a m
5
Feb 16, 2023
7:59 PM
= m a
6
Feb 16, 2023
7:59 PM
= m
1
+ m
2
7
Feb 16, 2023
8:00 PM
= F
foot
/( m
1
+ m
2
)
8
Feb 16, 2023
8:02 PM
= ( m
1
+ m
2
) g
9
Feb 16, 2023
8:02 PM
= - g ( m
1
+ m
2
)
10 Feb 16, 2023
8:02 PM
= - a ( m
1
+ m
2
)
11 Feb 16, 2023
9:18 PM
= F
foot
Part (b)
Choose the correct free-body diagram for the system made up of teenager and his skateboard. If appropriate, use
to represent the normal force, and use
to represent the combined weight of the teenager and the skateboard. Grade = 100%
Correct Answer
Student Final Submission
Feedback
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints Feedback
1
Feb 16, 2023
2:46 PM
Note: Feedback provided here but not accessed during assignment.
Which force allows the student to accelerate from rest.
2
Feb 16, 2023
2:46 PM
Note: Feedback provided here but not accessed during assignment.
Does the foot exert a force on the system comprised of the teenager and the skateboard? A
related force allows for the skateboard to accelerate.
3
Feb 16, 2023
2:46 PM
Note: Feedback provided here but not accessed during assignment.
Why does the student neither sink into the ground nor leave the ground when pushing off
with the foot?
4
Feb 16, 2023
2:46 PM
Note: Feedback provided here but not accessed during assignment.
Why does the student neither sink into the ground nor leave the ground when pushing off
with the foot? Does the foot exert a force on the system comprised of the teenager and the
skateboard?
5
Feb 16, 2023
2:46 PM
Note: Feedback provided here but not accessed during assignment.
Does the foot exert a force on the system comprised of the teenager and the skateboard?
6
Feb 16, 2023
2:46 PM
Part (c)
In terms of given quantities, write an expression for the magnitude of the skateboard's acceleration,
, while the teenager is pushing backwards on the ground.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
= F
foot
/( m
1
+ m
2
)
= F
foot
/( m
1
+ m
2
)
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
7:55 PM
= F
foot
/( m
1
+ m
2
)
Part (d)
What is the numerical value of the magnitude, in meters per squared second, of the acceleration?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
= = Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
7:56 PM
= Problem 15: Consider the block shown in the figure, which has a mass m
and is sitting
at rest on a ramp making an angle of θ
with respect to the floor. How big is the normal force acting on the block due to the ramp?
200.0
199.0
100.2
101.2
102.2
98.30
163.9
202.2
-
202.2
202.2
202.3
204.1
204.2
204.9
191.8
202.8
135.0
147.0
155.0
189.0
188.0
179.0
165.0
177.0
123.0
221.0
290.0
182.0
182.0
209.9
55.00
14.00
204.5
103.4
103.3
760.5
760.5
319.5
760.5
1.906
1.900
1.900
0.000
0.000
0.000
2.282
×
10
4
2.282
×
10
4
2.282
×
10
4
1.911
×
10
4
1.911
×
10
4
1.911
×
10
4
1.804
×
10
4
1.804
×
10
4
1.804
×
10
4
50.50
50.50
2.185
-
2.185
5.026
4.256
4.260
39.65
38.88
40.42
8.450
-
8.450
5.910
5.100
-
5.100
55.00
53.00
54.80
50.50
1382
1382
-
2764
2764
-
1382
1382
519.4
519.4
519.4
=
71.1
kg
m
1
=
15.2
N
F
foot
=
2.1
kg
m
2
F
ground
F
ground
F
ground
F
ground
F
ground
F
ground
F
ground
F
ground
F
ground
F
ground
F
ground
F
ground
F
ground
F
ground
F
N
F
g
a
a
a
a
a
0.2077 m/s
2
a
0.2100 m/s
2
a
0.2100 m/s
2
Grade = 100%
Correct Answer
Student Final Submission
Feedback
Less than mg
.
Less than mg
.
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints Feedback
1
Feb 16, 2023
2:45 PM
The answer depends on the
size of the frictional force.
Note: Feedback provided here but not accessed during assignment.
The frictional force is acting at a right angle with respect to the normal force,
and has no effect on the answer.
2
Feb 16, 2023
2:45 PM
Greater than mg
.
Note: Feedback provided here but not accessed during assignment.
Use Newton’s second law for the direction perpendicular to the incline. What
is the normal force in terms of the gravitational force and the angle of incline?
3
Feb 16, 2023
2:45 PM
Equal to mg sin(
θ
)
.
Note: Feedback provided here but not accessed during assignment.
When the angle is zero, the normal force should be equal to the gravitational
force.
4
Feb 16, 2023
2:45 PM
Equal to mg
.
Note: Feedback provided here but not accessed during assignment.
Use Newton’s second law for the direction perpendicular to the incline. What
is the normal force in terms of the gravitational force and the angle of incline?
5
Feb 16, 2023
2:45 PM
Less than mg
.
Problem 16: A block is resting on a wooden plank. There is a hinge on one end of the
plank which allows the other end to be lifted to create an angle, θ
, with respect to the
horizontal as shown in the figure. The coefficient of static friction between the block
and the plank is μ
s
. Part (a)
Please use the interactive area below to draw the Free Body Diagram for the block. Use Fs
to denote the force of static friction.
Grade = 0%
Give Up was accessed and a deduction for using the Give Up may have been applied.
Correct Answer
Student Final Submission
Feedback
It looks like there is at least
one force that is expected in
the drawing, but isn't in the
right direction. Grade Summary
Deduction for Final Submission
100
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
2
%
Student Grade = 100 - 100
- 2 = 0%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints Feedback
1
Feb 16, 2023
8:55 PM
Note: Feedback provided here but not
accessed during assignment.
It looks like there is at least one force that is
expected in the drawing, but isn't in the right
direction. 2
Feb 16, 2023
9:45 PM
Note: Feedback provided here but not
accessed during assignment.
It looks like there is at least one force that is
expected in the drawing, but isn't in the right
direction. 3
Feb 16, 2023
9:45 PM
Note: Feedback provided here but not
accessed during assignment.
It looks like there is at least one force that is
expected in the drawing, but isn't in the right
direction. 4
Feb 16, 2023
9:45 PM
Note: Feedback provided here but not
accessed during assignment.
It looks like there is at least one force that is
expected in the drawing, but isn't in the right
direction. 5
Feb 16, 2023
9:46 PM
Note: Feedback provided here but not
accessed during assignment.
It looks like there is at least one force that is
expected in the drawing, but isn't in the right
direction. 6
Feb 16, 2023
9:47 PM
It looks like there is at least one force that is
expected in the drawing, but isn't in the right
direction. Part (b)
The angle θ
is slowly increased. Write an expression for the angle at which the block begins to move in terms of μ
s
.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
θ
= atan(
μ
s
)
θ
= atan(
μ
s
)
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints Feedback
1
Feb 16, 2023
9:41 PM
θ
=
acotan(
μ
s
)
Note: Feedback provided here but not accessed during assignment.
Pay careful attention to trigonometric relationships and how they affect components of the
terms in your expression.
2
Feb 16, 2023
9:41 PM
θ
=
atan(
μ
s
)
Part (c)
If a student measures that the block begins to move at an angle of θ
= 23
°, what is the numerical value of the coefficient of static friction,
μ
s
?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
μ
s
= μ
s
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
9:41 PM
μ
s
= Problem 17: Consider a skier heading down a 10.5
° slope. Assume the coefficient of friction for waxed wood on wet snow is
μ
k
=0.10 and use a coordinate system in which down the slope is positive.
Randomized Variables
θ
= 10.5
°
Part (a)
Calculate the acceleration of the skier in m/s
2
.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
a
= a
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
9:01 PM
a
= Part (b)
Find the angle of the slope down which this skier could coast at a constant velocity in degrees.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
θ
= θ
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
9:01 PM
θ
= Problem 18: A bicycle and rider with mass m
rolls down a hill at constant velocity
under the influence of gravity with negligible rolling resistance. The hill makes an
angle of θ
= 11
degrees with respect to the horizontal. The force of air resistance,
termed "drag", has a magnitude of F
d
= 71
N. Assume that the x-direction is parallel to
the slope of the hill and the y-direction is perpendicular to the hill as shown. Randomized Variables
θ
= 11
°
F
d
= 71
N Part (a)
What is the sum of the forces in the y direction, Σ
F
y
in Newtons?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
Σ
F
y
= Σ
F
y
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
8:31 PM
Σ
F
y
= Part (b)
Input an expression for the sum of the forces in the x-direction, Σ
F
x
, in terms of the quantities given and variables available in the palette.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
Σ
F
x
= m g sin(
θ
) - F
d
Σ
F
x
= m g sin(
θ
) - F
d
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints Feedback
1
Feb 16, 2023
8:33 PM
Σ
F
x
= m g sin(
θ
)
Note: Feedback provided here but not accessed during assignment.
You have neglected to include the air friction.
2
Feb 16, 2023
8:33 PM
Σ
F
x
=
- g m sin(
θ
)
Note: Feedback provided here but not accessed during assignment.
It appears that there may be a sign error associated with at least one term in your
expression.
3
Feb 16, 2023
9:05 PM
Σ
F
x
=
- ( g ) ( m ) sin(
θ
)
Note: Feedback provided here but not accessed during assignment.
It appears that there may be a sign error associated with at least one term in your
expression.
4
Feb 16, 2023
9:05 PM
Σ
F
x
=
( m g )/sin(
θ
)
5
Feb 16, 2023
9:05 PM
Σ
F
x
=
( m g )/cos(
θ
)
6
Feb 16, 2023
9:05 PM
Σ
F
x
= m g cos(
θ
)
Note: Feedback provided here but not accessed during assignment.
Pay careful attention to trigonometric relationships and how they affect components
of the terms in your expression.
7
Feb 16, 2023
9:05 PM
Σ
F
x
=
- g m cos(
θ
)
Note: Feedback provided here but not accessed during assignment.
Pay careful attention to trigonometric relationships and how they affect components
of the terms in your expression.
It appears that there may be a sign error associated with at least one term in your
expression.
8
Feb 16, 2023
9:06 PM
Σ
F
x
= F
d
- g m
Note: Feedback provided here but not accessed during assignment.
You are missing the following parts for this term: sin(
θ
)
It appears that there may be a sign error associated with at least one term in your
expression.
9
Feb 16, 2023
9:07 PM
Σ
F
x
= F
d
+ g m
Note: Feedback provided here but not accessed during assignment.
You are missing the following parts for this term: sin(
θ
)
It appears that there may be a sign error associated with at least one term in your
expression.
10 Feb 16, 2023
9:07 PM
Σ
F
x
= F
d
+
α
m
Note: Feedback provided here but not accessed during assignment.
It appears that there may be a sign error associated with at least one term in your
expression.
11 Feb 16, 2023
9:07 PM
Σ
F
x
=
F
d
- g m sin(
θ
)
Note: Feedback provided here but not accessed during assignment.
You have confused positive and negative x-direction. Check closely the signs of all
components.
12 Feb 16, 2023
9:07 PM
Σ
F
x
=
F
d
+ g m sin(
θ
)
Note: Feedback provided here but not accessed during assignment.
Closely check signs of all components. Is the force of resistance directed in positive or
negative x-direction?
13 Feb 16, 2023
9:09 PM
Σ
F
x
=
m g sin(
θ
) - F
d
Part (c)
Using the information above, what is the total mass of the bike and rider m
in kg?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
m
= m
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
9:10 PM
m
= 2
Feb 16, 2023
9:11 PM
m
= Problem 19: A book with mass m
= 1.25
kg rests on the surface of a table. The
coefficient of static friction between the book and the table is μ
s
= 0.75
and the
coefficient of kinetic friction is μ
k
= 0.45
. Part (a)
Write an expression for F
m
the minimum force required to produce movement of the book on the surface of the table.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
F
m
= μ
s
m g
F
m
= μ
s
m g
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
8:18 PM
F
m
= μ
s
m g
Part (b)
Solve numerically for the magnitude of the force F
m
in Newtons.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
F
m
= F
m
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
8:19 PM
F
m
= Part (c)
Write an expression for a
, the book's acceleration, after it begins moving. (Assume the minimum force, F
m
, continues to be applied.)
Grade = 100%
Correct Answer
Student Final Submission
Feedback
a
= (
μ
s
-
μ
k
) g
a
= (
μ
s
-
μ
k
) g
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
8:20 PM
a
= (
μ
s
-
μ
k
) g
Part (d)
Solve numerically for the acceleration, a
in m/s
2
.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
a
= a
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
8:21 PM
a
= Problem 20: A man is attempting to lift a crate with a double-pulley system, as shown.
A single rope is routed from the ceiling, through the two pulleys, and to where the man
pulls on it. The crate has mass m
2
= 94
kg, and the man has m
1
= 75 kg. He pulls
downward on the rope with a force of magnitude F
= 618
N. The pulleys are both
massless and frictionless, and the rope is ideal. Part (a)
Let
be the tension in the rope. Enter an expression for the sum of the forces in the
direction acting on the crate.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
= 2 T - m
2
g
= 2 T - m
2
g
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
8:13 PM
= 2 T - m
2
g
Part (b)
Using the results from above, write an expression for the crate's vertical acceleration,
.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
= ( 2 T - m
2
g )/m
2
= ( ( 2 T )/( m
2
) ) - g
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints Feedback
1
Feb 16, 2023
8:14 PM
= ( 2 T )/( m
2
)
Note: Feedback provided here but not accessed during assignment.
You are missing the acceleration due to gravity. Reconsider your result from
part (a).
2
Feb 16, 2023
8:14 PM
=
( ( 2 T )/( m
2
) ) - g
Part (c)
What is the magnitude, in newtons, of the tension force,
?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
= = Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
8:15 PM
= 2
Feb 16, 2023
8:15 PM
= - 469.6
= 3
Feb 16, 2023
8:15 PM
= Part (d)
What is the acceleration, in meters per squared second, of the crate?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
= = Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
8:16 PM
= Problem 21: A block with mass m
1
= 8.1
kg rests on the surface of a horizontal table
which has a coefficient of kinetic friction of μ
k
= 0.61
. A second block with a mass m
2
= 8.3
kg is connected to the first by an ideal string passing over an ideal pulley such
that the second block is suspended vertically. The second block is released from rest,
and motion occurs. Part (a)
Using the variable T
to represent tension, write an expression for the sum of the forces in the y
-direction, Σ
F
y
, for block 2.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
Σ
F
y
= T - m
2
g
Σ
F
y
= T - m
2
g
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
8:06 PM
Σ
F
y
= T - m
2
g
Part (b)
Using the variable T
to represent tension, write an expression for the sum of the forces in the x
-direction, Σ
F
x
for block 1.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
Σ
F
x
= T -
μ
k
m
1
g
Σ
F
x
= T -
μ
k
m
1
g
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
8:06 PM
Σ
F
x
= T -
μ
k
m
1
g
Part (c)
Block 1 accelerates along the tabletop, in the horizontal direction, while block 2 moves vertically. With the coordinate system provided in
the drawing, we may write
and
. Write an expression that relates the vertical component of the acceleration of block 2 to the horizontal component of the acceleration of
block 1.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
a
2
= - a
1
a
2
= - a
1
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer Hints Feedback
1
Feb 16, 2023
8:07 PM
a
2
= a
1
Note: Feedback provided here but not accessed during assignment.
As block 1 accelerates towards the right, in the positive x
direction, block 2 accelerates
downward, in the negative y
direction.
2
Feb 16, 2023
8:07 PM
a
2
= - a
1
Part (d)
Write an expression using the variables provided for the magnitude of the tension force, T
.
Grade = 100%
Correct Answer
Student Final Submission
Feedback
T
= ( m
1
m
2
g (
μ
k
+ 1 ) )/( m
1
+ m
2
)
T
= ( m
1
m
2
g ( 1 +
μ
k
) )/( m
1
+ m
2
)
Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
8:08 PM
T
= ( m
1
m
2
g ( 1 +
μ
k
) )/( m
1
+ m
2
)
Part (e)
What is the tension, T
, in newtons?
Grade = 100%
Correct Answer
Student Final Submission
Feedback
T
= T
= Correct!
Grade Summary
Deduction for Final Submission
0
%
Deductions for Incorrect Submissions, Hints and Feedback [?]
0
%
Student Grade = 100 - 0
- 0 = 100%
Submission History
All Date times are displayed in Central Standard Time.
Red submission date times indicate late work.
Date
Time
Answer
Hints
Feedback
1
Feb 16, 2023
8:10 PM
T
= 15.778
T
= 2
Feb 16, 2023
8:11 PM
T
= All content © 2023 Expert TA, LLC
θ
Fn
Fs
Fg
Fg
Fs
Fn
Fg
Fn
Fs
θ
Fg
Fn
Fs
θ
Fg
Fn
Fs
θ
Fg
Fn
Fs
Fg
Fn
Fs
Fg
Fs
Fn
0.4245
0.4200
0.4200
0.8223
0.8200
0.8200
5.710
5.710
5.710
0.000
0.000
0.000
37.97
37.97
438.7
37.97
9.188
9.190
9.190
2.940
2.940
2.940
T
y
F
y
F
y
F
y
a
crate
a
crate
a
crate
a
crate
a
crate
T
T
618.0 N
T
618.0 N
T
469.6 N
T
T
-
469.6 N
T
618.0 N
a
crate
3.339 m/s
2
a
crate
3.350 m/s
2
a
crate
3.350 m/s
2
=
a
1
a
1
i
=
a
2
a
2
y
64.75
64.68
15.78
64.68
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help