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Serway and Vuille - College Physics 8/e (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 11 / 25

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

Last Saved : n/a Saving...  ()

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  • Instructions

    College Physics by Serway and Vuille includes all end-of-chapter problems, conceptual questions, quick quizzes, active figure simulations, and most worked examples. Questions provide answer-specific feedback, hints to guide students to content mastery and links to animated simulations and tutorial videos to further students' learning. A new Math Review feature allows instructors to assign essential mathematics examples to bring students up to speed.

    For select problems, 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, 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 problem, corresponding to Question 4.17 from Serway's text. This problem features conditional feedback for incorrect numerical answers and algorithmic solutions. You can see the solution by choosing the Practice another Version button. (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 problem, corresponding to Question 4.46 from Serway's text. This problem is an "Enhanced Content Problem" and is designed to help students see the relationship between physics principles, either verbally, symbolically or conceptually.

    Question 3 is an end-of-chapter problem, corresponding to Question 4.32 from Serway's text. This problem is a "Guided Problem", which break problems down into smaller steps so students see how concepts interrelate and build on each other.

    Question 4 is an Active Example, corresponding to Example 4.1 in Serway's text. By making the text's Worked Examples algorithmic and assignable, students will be guided through the process needed to master a concept, using the same style and format as the Examples in the text. Enhanced with robust feedback to help students avoid misconceptions and a host of useful tools including video and simulations, the Active Examples can be used as assignable mini-tutorials.

    Question 5 is a Multiple Choice Question, corresponding to Question 4.12 in Serway's text. These questions provide a key opportunity for students to check their conceptual understanding.

    Question 6 is an Active Figure question. Specific figures from the text have been recast as interactive animated simulations and made assignable so students can learn from observation. This question animates Figure 4.21 from Serway's text.

    Question 7 is a Math Review question from a special math review question bank, to test and review students' pre-existing mathematical skills.

    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. 2/4 points  |  Previous Answers SerCP8 4.P.017.soln. My Notes
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0/1 0/1 1/1 1/1
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2/4
 
A 660 N cat burglar is supported by cables in the figure below. Assume the angle θ of the inclined cable is 32.0°.


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.)
(a) Find the tension in each cable.
inclined cable Incorrect: Your answer is incorrect.
Your response differs from the correct answer by more than 10%. Double check your calculations. N
horizontal cable Incorrect: Your answer is incorrect.
Your response differs from the correct answer by more than 10%. Double check your calculations. N
vertical cable Correct: Your answer is correct. N

(b) Suppose the horizontal cable were reattached higher up on the wall. Would the tension in the inclined cable increase, decrease, or stay the same?
     Correct: Your answer is correct.


Note from WebAssign: All numerical end-of-chapter questions will give feedback to the student when an incorrect answer is submitted. Please see the algorithmically generated solution below.

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2. 2/2 points  |  Previous Answers SerCP8 4.P.046. My Notes
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1/1 1/1
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2/2
 
A hockey puck struck by a hockey stick is given an initial speed v0 in the positive x-direction. The coefficient of kinetic friction between the ice and the puck is μk.
(a) Obtain an expression for the acceleration of the puck. (Use the following as necessary: μk and g.)
a =
−μk​g
Correct: Your answer is correct.


(b) Use the result of part (a) to obtain an expression for the distance d the puck slides. The answer should be in terms of the variables v0, μk, and g only.
d =
(v0​)22μk​g​
Correct: Your answer is correct.

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3. 5/7 points  |  Previous Answers SerCP8 4.P.032. My Notes
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1/1 1/1 1/1 1/1 1/1 0/1 –/1
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5/7
 
Two blocks of masses m1 and m2 (m1 > m2) are placed on a frictionless table in contact with each other. A horizontal force of magnitude F is applied to the block of mass m1 in the figure below.
(a) If P is the magnitude of the contact force between the blocks, draw the free-body diagrams for each block. (Do this on paper. Your instructor may ask you to turn in this work.)

(b) What is the net force on the system on the system consisting of both blocks? (Use the following as necessary: P and F.)
Fnet =
F
Correct: Your answer is correct.


(c) What is the net force acting on m1? (Use the following as necessary: P and F.)
Fnet,1 =
F−P
Correct: Your answer is correct.


(d) What is the net force acting on m2? (Use the following as necessary: P and F.)
Fnet,2 =
P
Correct: Your answer is correct.


(e) Write the x-component of Newton's second law for each block. (Do this on paper. Your instructor may ask you to turn in this work.)

(f) Solve the resulting system of two equations and two unknowns, expressing the acceleration a and contact force P in terms of the masses and force. (Use the following as necessary: m1, m2, and F.)
a =
Fm1​+m2​​
Correct: Your answer is correct.
P =
m2​Fm1​+m2​​
Correct: Your answer is correct.

(g) How would the answers change if the force had been applied to m2 instead? (Hint: use symmetry; don't calculate!) (Do this on paper. Your instructor may ask you to turn in this work.)
(h) Is the contact force larger, smaller, or the same in this case?
    


(i) Why?

This answer has not been graded yet.

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4. 0/5 points  |  Previous Answers SerCP8 4.AE.01. My Notes
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0/5
 
EXAMPLE 4.1 Airboat
Goal Apply Newton's law in one dimension, together with the equations of kinematics.

Problem An airboat with mass 3.50 102 kg, including passengers, has an engine that produces a net horizontal force of 7.70 102 N, after accounting for forces of resistance.   (a)   Find the acceleration of the airboat.   (b)   Starting from rest, how long does it take the airboat to reach a speed of 12.0 m/s?   (c)   After reaching this speed, the pilot turns off the engine and drifts to a stop over a distance of 50.0 m. Find the resistance force, assuming it's constant.

Strategy In part (a), apply Newton's second law to find the acceleration, and in part (b) use this acceleration in the one-dimensional kinematics equation for the velocity. When the engine is turned off in part (c), only the resistance forces act on the boat, so their net acceleration can be found from v2 - v02 = 2aΔx. Then Newton's second law gives the resistance force.
SOLUTION
(a) Find the acceleration of the airboat.
Apply Newton's second law and solve for the acceleration.
ma = Fnet → a = Fnet = 7.70 102 N = 2.20 m/s2
m 3.50 102 kg
(b) Find the time necessary to reach a speed of 12.0 m/s.
Apply the kinematics velocity equation.
v = at + v0 = (2.20 m/s2)t = 12.0 m/s → t = 5.45 s
(c) Apply the kinematics velocity equation
Using kinematics, find the net acceleration due to resistance forces.
v2 - v02 = 2aΔx
0 - (12.0 m/s)2 = 2a(50.0 m) → a = -1.44 m/s2
Substitute the acceleration into Newton's second law, finding the resistance force:
Fresist = ma = (3.50 102 kg)(-1.44 m/s2) = -504 N
LEARN MORE
Remarks The propeller exerts a force on the air, pushing it backwards behind the boat. At the same time, the air exerts a force on the propeller and consequently on the airboat. Forces always come in pairs of this kind, which are formalized in the next section as Newton's third law of motion. The negative answer for the acceleration in part (c) means that the airboat is slowing down.

Question Choose the remaining forces acting on the airboat and its passenger, taken together. (Select all that apply.)
Incorrect: Your answer is incorrect.


Review Tip 4.2 in the textbook about which quantities are forces. Separately consider all the forces that act on the airboat in the horizontal direction, and all the forces that act in the horizontal direction. Which ones are already considered? Which forces are produced by gravity? Consider acceleration in the vertical direction and what that implies about the net vertical force on the boat. What force acts to produce that net vertical force despite the force of gravity?
PRACTICE IT
Use the worked example above to help you solve this problem. An airboat with mass 4.44 102 kg, including passengers, has an engine that produces a net horizontal force of 8.04 102 N, after accounting for forces of resistance.

(a) Find the acceleration of the airboat.
Incorrect: Your answer is incorrect.
Enter a number. m/s2

(b) Starting from rest, how long does it take the airboat to reach a speed of 15.4 m/s?
Incorrect: Your answer is incorrect.
Enter a number. s

(c) After reaching this speed, the pilot turns off the engine and drifts to a stop over a distance of 50.0 meters. Find the resistance force, assuming it's constant.
Incorrect: Your answer is incorrect.
Enter a number. N
EXERCISE HINTS:  GETTING STARTED  |  I'M STUCK!
Use the values from PRACTICE IT to help you work this exercise. Suppose the pilot, starting again from rest, opens the throttle only partway. At a constant acceleration, the airboat then covers a distance of 63.8 meters in 12.0 seconds. Find the net force acting on the boat.
Incorrect: Your answer is incorrect.
Enter a number. N

Note from WebAssign
If a student makes one of several common mistakes on the Master It 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
md/t2 197
(2md/t)2 2.23e+07
20md/t2 3930
md/(2t2) 98.4
Correct formula Correct answer
2md/t2 393

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5. –/3 points SerCP8 4.P.012. My Notes
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–/1 –/1 –/1
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–/3
 
Two forces are applied to a car in an effort to move it, as shown in the following figure, where F1 = 411 N and F2 = 366 N. (Assume up and to the right as positive directions.)
(a) What is the resultant of these two forces?
magnitude N
direction ° to the right of the forward direction

(b) If the car has a mass of 3,000 kg, what acceleration does it have? Ignore friction.
m/s2

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6. –/2 points was SerCP8 4.AF.21. My Notes
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Active Figure 4.21 - Acceleration Up and Down an Incline with Friction

Instructions: Click 'start' in the following active figure depicting acceleration up and down an incline with friction to complete the exercise.


Display in a New Window Explore
The following is a simple method of measuring coefficients of friction. Suppose a block is placed on a rough surface inclined relative to the horizontal as shown in the Active Figure. The incline angle is increased slowly until the block starts to move at a critical angle of θc = 22°. Use the measured value of the critical angle to find the coefficient of static friction μs.

Conceptualize
Consider the free-body diagram in the Active Figure. Consider the directions of the all the forces acting on the block, which are the gravitational force, the normal force, and the force of kinetic friction f.

Categorize
The critical angle is the largest angle for which the gravitational force does not yet cause the block to move. Finding this angle is a problem in which we deal with the block as a particle in equilibrium.

Analyze
Because the block is on an inclined surface, the normal force perpendicular to the surface does not balance the downward force of gravity. The normal force and the force of gravity produce a net force that tends to slide the block down the incline. Static friction acts along the direction of the surface to balance this combined force of gravity and the normal force. But as the angle of inclination increases, the combined force tending to slide the block increases, requiring a stronger force of static friction, until the force of static friction reaches its largest possible value given by

fs, max = μsn
At that point, any further increase in the angle of inclination results in unbalanced forces that accelerate the block. We need to find the largest possible angle of inclination θ for which the forces can still balance. We choose the x direction to be parallel to the inclined surface and the y perpendicular to it. Because the block is in equilibrium, the force components must balance separately along the x direction and along the y direction. The force of gravity in the x direction (mg) sin(θ) must be balanced by the force of static friction fs,
Equation (1) Σ Fx = mg sin(θ) - fs = 0
Similarly, along the y direction, the normal force vector n from the surface must balance the corresponding component of force from gravity:
Equation (2) Σ Fy = n - mg cos(θ) = 0
Solving Equation (2) for mg = n(cos(θ)) - 1 and substituting into Equation (1) gives
Equation (3) fs = mgsin(θ) = ( n ) sin(θ) = ntan(θ)
cos(θ)
When the block is on the verge of slipping, θ = θc and the force of static friction has reached its maximum value fs, max - μsn and θc into Equation (3) gives
Equation (4) μs = tan(θc)
For the measured critical angle of μc = 22°, we find from Equation (4) that
μs = tan( ) =

Finalize Once the block starts to move at θ > θc, it accelerates down the incline and the force of friction is fs = μkn. If θ is reduced to a value less than θc, however, it may be possible to find an angle θ such that the block moves down the incline with constant speed as a particle in equilibrium again (with acceleration a = 0). In this case, Equations (1) and (2) with fs replaced by fk lead to an expression that gives the value of the coefficient of kinetic fiction where θc, it is possible to find an angle θc such that the block moves down the incline with constant speed as a particle in equilibrium again (ax = 0). In this case, use Equations (1) and (2) with fx replaced by fk to find μk = tan(θc where θc < θc.

Hint

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Note from WebAssign: Each Serway textbook includes a Math Review Appendix. One such question is below. For an assignment of Math Review questions click here.

Quadratic Equations

Solve the following quadratic equation.
x2 + 11x - 12 = 0
x = Correct: Your answer is correct. (smaller value)
x = Correct: Your answer is correct. (larger value)

Refer to th.
Appendix A.3D

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