WebAssign is not supported for this browser version. Some features or content might not work. System requirements

WebAssign

Welcome, demo@demo

(sign out)

Tuesday, April 1, 2025 05:23 EDT

Home My Assignments Grades Communication Calendar My eBooks

WebAssign - University Physics 1/e (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 0 / 26

Due : Monday, January 28, 2030 00:00 EST

Last Saved : n/a Saving...  ()

Print Assignment
Question
Points
1 2 3 4 5 6 7 8 9 10
0/1 0/2 0/2 –/3 –/5 –/3 0/3 –/1 –/3 –/3
Total
0/26 (0.0%)

  • Instructions

    University Physics is our new premiere collection of calculus-based physics content, available to be used with any textbook (or no textbook at all!).

    University Physics is authored by experienced physics instructors with years of teaching the introductory physics sequence, the new University Physics question collection by WebAssign includes over 1800 questions covering every concept in the calculus-based course, designed to work with any textbook (or no textbook at all).

    WebAssign Content for University Physics includes:
    • A collection of over 1800 questions representing every concept covered in the calculus-based course
    • Answer feedback for every question and question part, designed specifically to address student misconceptions
    • Multi-part tutorials for key concepts that reinforce problem-solving skills and conceptual understanding
    • Detailed stepped-out solutions for every question, available at the instructors discretion
    • A Math Personal Study Plan to strengthen student prerequisite skills
    This demo assignment allows many submissions and allows you to try another version of the same question for practice wherever the problem has randomized values.

    The answer key and solutions will display after the first submission for demonstration purposes. Instructors can configure these to display after the due date or after a specified number of submissions.

Assignment Submission

For this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer.

Assignment Scoring

Your last submission is used for your score.

1. 0/1 points  |  Previous Answers WebAssignCalcPhys1 4.P.015.Tutorial. My Notes
Question Part
Points
Submissions Used
1
0/1
1/50
Total
0/1
 
You are walking around your neighborhood and you see a child on top of a roof of a building kick a soccer ball. The soccer ball is kicked at 51° from the edge of the building with an initial velocity of 17 m/s and lands 63 meters away from the wall. How tall is the building that the child is standing on?
Incorrect: Your answer is incorrect. seenKey

92.1
What information is given, and what is unknown? What will you need to know to find the height of the building? How can you find the time of flight for the ball? m




Solution or Explanation
First state the givens:
x0 = 0
x = 63 m
y = 0
v0 = 17 m/s
θ = 51°.

Now we can find the velocity components:
v0x = v0 cos θ
 = (17 m/s)cos(51°)
 = 10.7 m/s
v0y = v0 sin θ
 = (17 m/s)sin(51°)
 = 13.2 m/s.

Using the x component of velocity we can find the time it takes for the ball to travel through the air:
x = x0 + v0x Δt
x = 0 + v0x Δt
Δt = 
x
v0x
 
Δt = 
63 m
10.7 m/s
 = 5.89 s.

Given the time of flight we can now find the height of the building. Consider the ground level to be
y = 0,
 and the positive y direction to be upwards.
y = y0 + v0yΔt + 
1
2
ayΔt2
0 = y0 + v0yΔt − 
1
2
gΔt2
y0 = −v0yΔt + 
1
2
gΔt2
y0 = −(13.2 m/s)(5.89 s) + 
1
2
(9.8 m/s2)(5.89 s)2
y0 = 92.1 m

The height of the building is 92.1 m.

Supporting Materials

Additional Materials

Your work in question(s) will also be submitted or saved.
Viewing Saved Work Revert to Last Response
2. 0/2 points  |  Previous Answers WebAssignCalcPhys1 5.P.024.Tutorial. My Notes
Question Part
Points
Submissions Used
1 2
0/1 0/1
1/50 1/50
Total
0/2
 
A contestant in a winter games event pushes a 36.0-kg block of ice across a frozen lake with a force of 25 N at 30.0° below the horizontal as shown in Figure (a) below, and it moves with an acceleration of 0.600 m/s2 to the right.
(a) What is the normal force exerted by the lake surface on the block of ice?
Incorrect: Your answer is incorrect. seenKey

365
Did you draw a free-body diagram and identify the forces acting on the system? Did you consider all applicable forces (or component of forces) in the vertical direction? N

(b) Instead of pushing on the block of ice, the contestant now pulls on it with a rope at the same angle above the horizontal as in part (a), and with the same magnitude of force. See Figure (b) above. Now what is the normal force exerted by the lake surface on the block of ice?
Incorrect: Your answer is incorrect. seenKey

340
Consider the change in the direction of the applied force in this case. N


Solution or Explanation
(a) The free-body diagram for the system (the block of ice) is shown below. Here
N
 is the normal force exerted by the lake surface on the block of ice. Note that the applied force vector
F
 is drawn with its tail at the system to easily visualize its components.
WebAssign Plot
Since we are interested in determining the magnitude of the normal force, we need to consider motion in the vertical direction when applying Newton's second law. Note that the acceleration in the vertical direction is zero as there is no motion in that direction.

Newton's second law for the vertical direction can therefore be written as follows.
N − mg − F sin θ = 0
N = mg + F sin θ
 = (36.0 kg)(9.80 m/s2) + (25.0 N)sin(30.0°)
 = 365 N

(b) In this case, the pulling force is pointing up and to the right. The sine component of
F
 is therefore directed upward now.
WebAssign Plot
As in part (a), the acceleration in the vertical direction is zero and Newton's second law can be written as follows.
N − mg + F sin θ = 0
N = mg − F sin θ
 = (36.0 kg)(9.80 m/s2) − (25.0 N)sin(30.0°)
 = 340 N

Supporting Materials

Additional Materials

Your work in question(s) will also be submitted or saved.
Viewing Saved Work Revert to Last Response
3. 0/2 points  |  Previous Answers WebAssignCalcPhys1 5.P.030. My Notes
Question Part
Points
Submissions Used
1 2
0/1 –/1
1/50 0/50
Total
0/2
 
Tom enlists the help of his friend John to move his car. They apply forces to the car as shown in the diagram. Here
|F1| = 442 N,
|F2| = 330 N
 and friction is negligible. Mass of the car = 3.50 103 kg, θ1 = 12.0°, and θ2 = 25.0°. The diagram below shows the top view of the car which is in the x-z plane (horizontal plane).

(a) Find the resultant force exerted on the car. (Express your answer in vector notation.)
Fnet
 =
√2
Incorrect: Your answer is incorrect. webMathematica generated answer key
Did you consider the components of the forces in the x and z directions? Review rules for vector addition. N

(b) What is the acceleration of the car? (Express your answer in vector notation.)
a
 =
m/s2

Supporting Materials

Your work in question(s) will also be submitted or saved.
Viewing Saved Work Revert to Last Response
4. –/3 points WebAssignCalcPhys1 5.P.076.Tutorial. My Notes
Question Part
Points
Submissions Used
1 2 3
–/1 –/1 –/1
0/50 0/50 0/50
Total
–/3
 
Alex is asked to move two boxes of books in contact with each other and resting on a rough floor. He decides to move them at the same time by pushing on box A with a horizontal pushing force FP = 9.5 N. Here, A has a mass mA = 10.2 kg and B has a mass mB = 7.0 kg. The contact force between the two boxes is
FC.
 The coefficient of kinetic friction between the boxes and the floor is 0.02. (Assume
FP
 acts in the +x direction.)
(a) What is the magnitude of the acceleration of the two boxes?
m/s2

(b) What is the force exerted on mB by mA? In other words, what is the magnitude of the contact force
FC?
N

(c) If Alex were to push from the other side on the 7.0-kg box, what would the new magnitude of
FC
 be?
N

Supporting Materials

Additional Materials

Your work in question(s) will also be submitted or saved.
Viewing Saved Work Revert to Last Response
5. –/5 points WebAssignCalcPhys1 6.P.012. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5
–/1 –/1 –/1 –/1 –/1
0/50 0/50 0/50 0/50 0/50
Total
–/5
 
A conical pendulum is a weight or bob fixed on the end of a string suspended from a pivot. It moves in a horizontal circular path, as shown in the diagram below.
(a) Write an expression for tension in the string. (Use the following as necessary: m, g, and θ.)
T =


(b) Write an expression for the centripetal acceleration of the bob. (Use the following as necessary: g, and θ.)
ac =


(c) Suppose the bob has a mass of 0.35 kg, the length of the pendulum is 0.75 m and the angle that it swings at is 14°. What is the magnitude of the centripetal acceleration?
m/s2

(d) What is the radius of the horizontal circular path?
m

(e) What is the speed of the mass?
m/s

Supporting Materials

Your work in question(s) will also be submitted or saved.
Viewing Saved Work Revert to Last Response
6. –/3 points WebAssignCalcPhys1 6.P.027.Tutorial. My Notes
Question Part
Points
Submissions Used
1 2 3
–/1 –/1 –/1
0/50 0/50 0/50
Total
–/3
 
The mass of a roller coaster car, including the riders, is 494 kg. The path of a portion of the ride is shown in the diagram below. The radii of curvature at the three points A, B, and C are 18.8 m, 11.6 m, and 6.80 m, respectively. The car, which travels from right to left, has a speed of 12.8 m/s and 22.5 m/s at A and B, respectively.
(a) What is the force exerted by the track at these two locations? (Enter the magnitudes of the forces.)
At A     N
At B     N

(b) At what minimum speed will the car lose contact with the track at C?
m/s

Supporting Materials

Additional Materials

Your work in question(s) will also be submitted or saved.
Viewing Saved Work Revert to Last Response
7. 0/3 points  |  Previous Answers WebAssignCalcPhys1 7.P.032. My Notes
Question Part
Points
Submissions Used
1 2 3
0/1 0/1 0/1
1/50 1/50 1/50
Total
0/3
 
An object starts from rest at the origin and moves along the x axis subject to the force shown in the figure below. If the mass of the object is 64.0 kg, determine the speed of the object at the following positions. Each unit along the position axis is 4.00 m and each unit along the force axis is 3.00 N.
WebAssign Plot
(a)    x = A
Incorrect: Your answer is incorrect. seenKey

1.3
How can you determine the net work a force does on an object over some displacement from a plot of the force versus the displacement? If you know the initial kinetic energy of an object and the net work done on the object for some displacement, how can you determine the final kinetic energy, and hence the final speed of the object? m/s

(b)    x = B
Incorrect: Your answer is incorrect. seenKey

2.25
How can you determine the net work a force does on an object over some displacement from a plot of the force versus the displacement? If you know the initial kinetic energy of an object and the net work done on the object for some displacement, how can you determine the final kinetic energy, and hence the final speed of the object? m/s

(c)    x = C
Incorrect: Your answer is incorrect. seenKey

2.6
How can you determine the net work a force does on an object over some displacement from a plot of the force versus the displacement? If you know the initial kinetic energy of an object and the net work done on the object for some displacement, how can you determine the final kinetic energy, and hence the final speed of the object? m/s


Solution or Explanation
(a) If an object is moving along the x axis under the action of a variable force
Fx,
 and we construct a plot of
Fx
 versus x, then the area bounded by the
Fx
 curve and the x axis for some displacement is the work done on the object by the force for this displacement and may be written as
area under curve = 
FxΔx = W.
According to the work-kinetic energy theorem, the net work done on an object is its change in kinetic energy. This may be written as
Wnet = ΔK = Kf − Ki.
For part (a), as the object moves from
x = 0
 to
x = 12.0 m,
 we have
Ki = 0,
 because the object is initially at rest.
Wnet = area under curve = 
(base)(height)
2
 = 
(12.0 m)(9.00 N)
2
 = 54.0 J
Kf = 
mvf2
2
 = Wnet, or vf = 
2Wnet
m
Entering values, we have
vf = 
2(54.0 J)
64.0 kg
 = 1.30 m/s.

(b) For part (b), as the object moves from
x = 12.0 m
 to
x = 24.0 m,
 we have
Ki = 54.0 J,
 because the final kinetic energy from part (a) is the initial kinetic energy for part (b).
Wnet = area under curve = (base)(height) = (12.0 m)(9.00 N) = 108 J
Kf = 
mvf2
2
 = Wnet + Ki, or vf = 
2(Wnet + Ki)
m
Entering values, we have
vf = 
2(108 J + 54.0 J)
64.0 kg
 = 2.25 m/s.

(c) For part (c), as the object moves from
x = 24.0 m
 to
x = 36.0 m,
 we have
Ki = 162 J,
 because the final kinetic energy from part (b) is the initial kinetic energy for part (c).
Wnet = area under curve = 
(base)(height)
2
 = 
(9.00 m)(12.0 N)
2
 = 54.0 J
Kf = 
mvf2
2
 = Wnet + Ki, or vf = 
2(Wnet + Ki)
m
Entering values, we have
vf = 
2(54.0 J + 162 J)
64.0 kg
 = 2.60 m/s.

Supporting Materials

Your work in question(s) will also be submitted or saved.
Viewing Saved Work Revert to Last Response
8. –/1 points WebAssignCalcPhys1 8.P.025. My Notes
Question Part
Points
Submissions Used
1
–/1
0/50
Total
–/1
 
As shown in the figure below, three blocks with masses m1 = 3.40 kg, m2 = 12.0 kg, and m3 = 19.0 kg, respectively, are attached by strings over frictionless pulleys. The horizontal surface exerts a 32.0-N force of friction on m2. If the system is released from rest, use energy concepts to find the speed of m3 after it moves down 3.00 m.
m/s

Supporting Materials

Your work in question(s) will also be submitted or saved.
Viewing Saved Work Revert to Last Response
9. –/3 points WebAssignCalcPhys1 9.P.053. My Notes
Question Part
Points
Submissions Used
1 2 3
–/1 –/1 –/1
0/50 0/50 0/50
Total
–/3
 
Ernest Rutherford (the first New Zealander to be awarded the Nobel Prize in chemistry) demonstrated that nuclei were very small and dense by scattering helium-4 nuclei (4He) from gold-197 (197Au). See the figure below.
The energy of the incoming helium nucleus was 6.99 10-13 J, and the masses of the helium and gold nuclei were 6.68 10-27 and 3.29 10-25 kg, respectively (note that their mass ratio is 4 to 197). If a helium nucleus scatters to an angle of 120° during an elastic collision with a gold nucleus, calculate the helium nucleus's final speed. (Enter your answer to at least three significant figures.)
m/s

Calculate the final velocity (magnitude and direction) of the gold nucleus. (Assume the positive x direction is the direction in which the helium nucleus is initially traveling, and that it scatters 120° clockwise from the +x axis.)
magnitude      m/s
direction     ° counterclockwise from the +x axis

Supporting Materials

Your work in question(s) will also be submitted or saved.
Viewing Saved Work Revert to Last Response
10. –/3 points WebAssignCalcPhys1 11.P.037. My Notes
Question Part
Points
Submissions Used
1 2 3
–/1 –/1 –/1
0/50 0/50 0/50
Total
–/3
 
The figure shows an overhead view of a 1.90-kg plastic rod of length 1.20 m on a table. One end of the rod is attached to the table, and the rod is free to pivot about this point without friction. A disk of mass 33.0 g slides toward the opposite end of the rod with an initial velocity of 25.0 m/s. The disk strikes the rod and sticks to it. After the collision, the rod rotates about the pivot point.
(a) What is the angular velocity of the two after the collision?
rad/s

(b) What is the kinetic energy before and after the collision?
KEi
 =		 J
KEf
 =		 J

Supporting Materials

Your work in question(s) will also be submitted or saved.
Viewing Saved Work Revert to Last Response
Home My Assignments
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.
Enter a number.