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 05:30 EDT

Home My Assignments Grades Communication Calendar My eBooks

Serway and Jewett - Physics for S&E 6/e (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 7 / 28

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

Last Saved : n/a Saving...  ()

Print Assignment
Question
Points
1 2 3 4 5
1/3 4/5 1/10 0/8 1/2
Total
7/28 (25.0%)

  • Instructions

    Physics for Scientists and Engineers 6/e by Serway and Jewett now has animations with conceptual questions and tutorial problems offering feedback and hints to guide student content mastery, as well as over 1600 end-of-chapter questions.

    Students will see a detailed solution, in blue, using the algorithmically-generated values in the question, at the instructor's discretion--for example, after a particular submission or after the assignment due date has passed.

    You can check out a sampling of this exciting development below. WebAssign is the most utilized homework system in physics. Designed by physicists for physicists, WebAssign, easy to use, reliable--a trusted companion to your teaching. Sign up for a Test Drive today!

    Question 1 is a traditional end-of-chapter question, with conditional feedback for incorrect numerical answers. (For this demo, the solution is displayed immediately after the question has been submitted using the algorithmically-generated values.)

    Question 2 is an end-of-chapter question enhanced by adding a link to an Active Figure. The solution is displayed after the question has been answered.

    Question 3 shows an Active Figure question which has been imported into WebAssign.

    Question 4 is a new Active Example which helps guide students through the process needed to master a concept.

    Question 5 is a Quick Quiz question which provides students opportunities to test their understanding of the physical concepts just presented.

    Click here for a list of all of the questions coded in WebAssign. This demo assignment allows many submissions and allows you to try another version of the same question for practice.

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. 1/3 points  |  Previous Answers PSE6 5.P.021. My Notes
Question Part
Points
Submissions Used
1 2 3
0/1 0/1 1/1
11/50 5/50 3/50
Total
1/3
 
The systems shown in Figure P5.21 are in equilibrium with m = 6.90 kg and θ = 25.0°. If the spring scales are calibrated in newtons, what do they read? (Ignore the masses of the pulleys and strings, and assume the incline is frictionless.)
scale in (a) Incorrect: Your answer is incorrect. N Your response differs from the correct answer by more than 100%.
scale in (b) Incorrect: Your answer is incorrect. N Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error.
scale in (c) Correct: Your answer is correct. N

Figure P5.21
Your work in question(s) will also be submitted or saved.
Viewing Saved Work Revert to Last Response
2. 4/5 points  |  Previous Answers was PSE6 5.P.028.AF. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5
0/1 1/1 1/1 1/1 1/1
7/50 3/50 3/50 2/50 2/50
Total
4/5
 
Two objects with masses of 2.60 kg and 5.00 kg are connected by a light string that passes over a light frictionless pulley to form an Atwood machine as shown in Figure 5.14a.

Figure 5.14a
(a) Determine the tension in the string.
Incorrect: Your answer is incorrect. N Your response differs from the correct answer by more than 10%. Double check your calculations.

(b) Determine the acceleration of each object.
2.60 kg mass Correct: Your answer is correct. m/s2 upwards
5.00 kg mass Correct: Your answer is correct. m/s2 downwards
(c) Determine the distance each object will move in the first second of motion if they start from rest.
2.60 kg mass Correct: Your answer is correct. m
5.00 kg mass Correct: Your answer is correct. m

Refer to the Active Figure Simulation to review the concepts this question addresses. Hint: Active Figure 5.14
Your work in question(s) will also be submitted or saved.
Viewing Saved Work Revert to Last Response
3. 1/10 points  |  Previous Answers was PSE6 5.AF.12. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5 6 7 8 9 10
–/1 –/1 0/1 0/1 –/1 1/1 0/1 –/1 –/1 –/1
0/50 0/50 1/50 1/50 0/50 1/50 1/50 0/50 0/50 0/50
Total
1/10
 
Active Figure 5.12 Pushing Blocks With A Constant Force

In the simulation below, you can observe the effects of a force applied to a two-block system.

Instructions: Use the direction toggle (dir'n) to set the direction of the external force, and click start to apply the force.
Newton's Second Law

Exploration 1
Push the direction toggle so the direction arrow points in the positive direction. Is this direction to your left or to the right? Also, before proceeding, reset the positions of the two masses to zero.
  1. Start the simulation and observe the masses, as they accelerate to the right on a frictionless surface and eventually disappear off the screen.
  2. Observe the on-screen values of time, distance and force and notice that the force remains constant throughout the motion.
  3. Record the final values of distance and force at the finishing time of 10 seconds.
Exploration 2
  1. As you observed in Exploration 1, the external force applied by the hand is constant at all times. Does this imply that the acceleration is constant, getting less, or increasing during the motion sequence?
  2. Note that the two masses are at rest at time zero. Use the final values of time and distance to calculate the acceleration in m/s2, without using the on-screen value of the force. In other words, use the kinematic equations to calculate the acceleration.
  3. Reverse the direction with the toggle, and find the acceleration without using the on-screen value of the force. Take particular care with the signs in calculating this acceleration.


Display in a New Window

Exploration 3
Use Newton's second law to confirm that the force applied by the hand is +0.1 N in the positive direction, and -0.1 N for motion in the negative direction.

Summary
(a) What is the acceleration magnitude for the two blocks in this simulation?
m/s2

Newton's Third Law and Contact Forces

According to Newton's third law, isolated forces cannot exist: forces always occur in action-reaction pairs. The contact forces between the two blocks in this simulation are an excellent example of action-reaction pairs.

Applied Force to the Right

Set the simulation so that the external force applied by the hand acts in the positive direction. In this case, the external force is applied to mass m1.

The free-body diagram for mass m1 is shown below.

Thus, m1 is acted on by two horizontal forces: the external force F to the right, and a contact force P21 acting to the left. Newton's second law for mass m1 gives:
Fx = F - P21 = m1a
(b) The applied force F must make both blocks accelerate. Solve this expression for the contact force acting on mass m1.
P21 = N
Now consider mass m2.

The only horizontal force acting on this mass is the contact force from m1: P12. Thus, Fx = P12 = m2a.

(c) Solve this for the contact force acting on m2:
P12 = Incorrect: Your answer is incorrect. N Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.
Notice that P21 points to the left while P12 points to the right -- so they really are different forces. Nevertheless, to within round-off errors, you should find that their magnitudes are equal.

Applied Force to the Left

Set the simulation so that the external force applied by the hand acts in the negative direction. In this case, the external force is applied to mass m2.

(d) Isolate mass m2, apply Newton's second law, and solve for the contact force.
P12 = Incorrect: Your answer is incorrect. N Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully.
(e) Isolate m1, apply Newton's second law, and solve for the contact force acting on m1.
P21 = N
Note that the contact force is now greater, since it must make the heavier block accelerate.


Exercise 5.12

Suppose that the masses in the simulation are changed so that m1 = 10.8 kg and m2 = 6.7 kg. Suppose also that a new external force F is applied so that the acceleration of the two blocks is 1.35 m/s2.

(f) What is the magnitude of the applied force?
F = Correct: Your answer is correct. N
(g) If the external force is applied in the positive direction, so that it acts on m1, find the magnitude of the contact forces.
P21 = Incorrect: Your answer is incorrect. N Your response differs from the correct answer by more than 100%.
P12 = N
(h) If the external force is applied in the negative direction, so that it acts on m2, find the magnitude of the contact forces.
P21 = N
P12 = N
Hint
Your work in question(s) will also be submitted or saved.
Viewing Saved Work Revert to Last Response
4. 0/8 points  |  Previous Answers PSE6 5.AE.01. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5 6 7 8
–/1 –/1 –/1 –/1 –/1 –/1 0/1 –/1
0/50 0/50 0/50 0/50 0/50 0/50 2/50 0/50
Total
0/8
 
Example 5.1 An Accelerating Hockey Puck
Problem A 0.30 kg hockey puck slides on the horizontal frictionless surface of an ice rink. It is struck simultaneously by two different hockey sticks. The two constant forces that act on the puck as a result of the hockey sticks are parallel to the ice surface and are shown in the pictorial representation in the figure. The force 1 has a magnitude of 5.6 N, and 2 has a magnitude of 8.0 N. If 1 = 23° and 2 = 64°, determine the acceleration of the puck while it is in contact with the two sticks.

Strategy Model the puck as a particle under a net force. Find the components of the force in each direction, then use Newton's second law to find the corresponding components of the acceleration. Resolve these components using trigonometry into the magnitude and direction of the acceleration.
Figure 4.4 from book
Figure 4.4 A hockey puck moving on a frictionless surface accelerates in the direction of the net force, = 1 + 2.

Solution
We first find the components of the net force. The component of the net force in the x direction is:
Fx = F1x + F2x = F1 cos 23° + F2 cos 64° = N
Next we find the component of the net force in the y direction.
Fy = F1y + F2y = -F1 sin 23° + F2 sin 64° = N
Now we use Newton's second law in component form to find the x and y components of acceleration:
ax = (Sigma F_x)/m = m/s2

ay = (Sigma F_y)/m = m/s2
The acceleration has the following magnitude.
a = sqrt(a_x^2 + a_y^2) = m/s2
Its direction is calculated relative to the positive x axis.
= tan-1 \(a_y / a_x\) = °
Decorative only

Decorative only
Exercise 5.1
Decorative only
Suppose three hockey sticks strike the puck simultaneously, with two of them exerting the forces from the example above. The result of the three forces is that the hockey puck shows no acceleration. What must be the components of the third force?
F3x = Incorrect: Your answer is incorrect. NYour response differs from the correct answer by more than 10%. Double check your calculations.
F3y = N
 Decorative only
Decorative only

Note from WebAssign
If a student makes one of several common mistakes on the Exercise portion above, they will get feedback specific to the mistake they made. If their answer is incorrect but does not meet one of these specific conditions, they will still get generic numerical feedback as to how far off they are, or if they appear to have made just a sign or order of magnitude error, etc.

Try some of these values to see what feedback you get!
  Incorrect formulaIncorrect answer
F3x -Fx' where Fx' is an incorrect answer entered for the first answer blank, above. ?
+Fx 9.82
F3y -Fy' where Fy' is an incorrect answer entered for the second answer blank, above. ?
+Fy -1.91
  Correct formula Correct answer
F3x -Fx -9.82
F3y -Fy 1.91
Your work in question(s) will also be submitted or saved.
Viewing Saved Work Revert to Last Response
5. 1/2 points  |  Previous Answers PSE6 5.QQ.06. My Notes
Question Part
Points
Submissions Used
1 2
1/1 –/1
4/50 0/50
Total
1/2
 
Suppose you are talking by interplanetary telephone to your friend, who lives on the Moon. He tells you that he has just won a newton of gold in a contest. Excitedly, you tell him that you entered the Earth version of the same contest and also won a newton of gold! Who is richer?
     Correct: Your answer is correct.

Explain.

This answer has not been graded yet.

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.
Enter a number.
Enter a number.
Enter a number.
Enter a number.