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Serway and Vuille - College Physics 10/e (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 7 / 53

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

Last Saved : n/a Saving...  ()

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  • Instructions

    The EWA course for the new edition for College Physics 10/e by Raymond A. Serway and Chris Vuille is more robust than ever to enhance your teaching experience and help you elevate students’ thinking. All end-of-chapter problems, Worked Examples, and Quick Quizzes are available in the EWA course, along with Active Figure questions and tutorial problems offering feedback and hints to guide students to content mastery.

    What's New in this Edition?
    • Warm-Up Exercises to help students review mathematical and physical concepts that are prerequisites for a given chapter’s problems set.
    • Training Tutorials, based on concepts from the text, guide students through how to solve a problem.
    • Algorithmic worked-out solutions are now available for quantitative end-of-chapter problems that do not include Master Its or Watch Its, ensuring that students receive some kind of guidance to help them solve the problem.
    • Pre-Lecture Explorations that introduce students to a concept through an interactive simulation and guided questions.

    In this assignment we present several textbook question types found in College Physics 10/e.

    Question 1 is a Warm-Up Exercise.

    Question 2 is a Training Tutorial that offers students another training tool to assist them in understanding how to apply certain key concepts presented in a given chapter.

    Question 3 is a Pre-Lecture Exploration. It combines an Active Figure with conceptual and analytical questions that guide students to a deeper understanding and help promote a robust physical intuition.

    Question 4 is an Active Example which guides students through the process needed to master a concept. A new "Review" question at the end provides a twist on the in-text Example to test student understanding.

    Question 5 is a Conceptual Question which serves as a concept check to help students test their understanding of physical concepts as they work through each chapter.

    Question 6 and 7 are traditional end-of-chapter problems with symbolic answer entry.

    Question 8 is a Master It problem with a complete tutorial.

    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 wherever the problem has randomized values.

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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.

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1. 2/2 points  |  Previous Answers SerCP10 4.WU.003. My Notes
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A force of 33.0 N is applied in the positive x-direction to a block of mass 8.05 kg, at rest on a frictionless surface. (Indicate the direction with the sign of your answer.)
(a) What is the block's acceleration?
Correct: Your answer is correct. m/s2

(b) How fast is it going after 6.50 s?
Correct: Your answer is correct. m/s
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2. 5/11 points  |  Previous Answers SerCP10 4.T.002. My Notes
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Tutorial 4.2: Applying the Second Law to Objects in Equilibrium Newton's second law is described by the equation
F
 = ma.
 An object is said to be in equilibrium when it isn't accelerating. In these cases,
a = 0
 and
F
 = 0.
This is a vector equation and is equivalent to the two-dimensional set of component equations given by
Fx = 0 and
 
Fy = 0.
The Tutorial Problems that follow focus on applying Newton's second law to objects in equilibrium and guide you through the Problem-Solving Strategy.
•   Read the problem carefully at least once.
•   Draw a picture of the system, identify the object of interest, and indicate forces with arrows.
•   Label each force in the picture.
•   Draw a free-body diagram of the object of interest.
•   Apply Newton's second law.
•   Solve for the desired unknown quantity, and substitute the numbers.
Guided Problem
The blocks shown in the figure have masses
m1 = 5.30 kg
 and
m2 = 2.70 kg,
 and are hanging in equilibrium from two cords of negligible mass. Determine the magnitude of the tension in the top cord.
Part 1 of 8
Read the problem carefully at least once.
Be sure to notice quantities that are known and those we're being asked to find. In this case, the masses are known and we're asked to find the magnitude of the tension force in the top cord.
Part 2 of 8
Draw a picture of the system, identify the object of interest, and indicate forces with arrows.
The figure below shows a picture of the blocks (our objects of interest).
Does the figure show all the forces acting on the blocks?
     Correct: Your answer is correct.
Correct. The figure shows all forces acting on the blocks.
Label each force in the picture.
Be sure each force in the figure is labeled in a way that brings to mind the associated physical quantity. The magnitudes of the tensions in the top and bottom cords are
T1
 and
T2,
 respectively. Similarly, the magnitudes for the forces of gravity are
Fg1
 and
Fg2.
Part 3 of 8
Draw a free-body diagram (FBD) of the objects of interest.

Which of the following free-body diagrams is correct?
     Incorrect: Your answer is incorrect.

Incorrect. Make sure each force is labeled so that you can identify it. The forces in this FBD are not labeled.
Part 4 of 8
Apply Newton's second law.
Applying Newton's second law means that we should take
F
 = 0
 for an object in equilibrium and write out each component. In this case, we'll take the positive y-direction to be straight up. Because there are no horizontal forces in this problem, only the y-components are relevant here.

Consider the correct answer above for the FBDs describing the blocks. Which of the following is the correct representation of the y-component of Newton's second law applied to
m2?
 Recall that
T2
 represents the magnitude of the tension force in the bottom cord and
Fg2
 represents the magnitude of the gravity force on
m2.
     Correct: Your answer is correct.
The correct form for Newton's second law applied to m2 is
T2 − Fg2 = 0.
Part 5 of 8
Apply Newton's second law (cont.).
The correct form for Newton's second law applied to m2 is
T2 − Fg2 = 0.
 Now let's find out whether or not any of the quantities in this expression are known.

Is
Fg2
 a known quantity?
     Correct: Your answer is correct.
Correct. The quantity
Fg2
 is the magnitude of the lower block's weight. The magnitude of the weight of any object is given by
Fg = mg.
 Because we know both
m2
 and
g,
Fg2
 is known.
Part 6 of 8
Apply Newton's second law (cont.).
Substituting
Fg2 = m2g
 into Newton's second law gives
T2 − Fg2 = T2 − m2g = 0,
and solving for the tension
T2
 in the lower cord, we have
T2 = m2g.
Let's continue to use Newton's second law by applying it to
m1.

Looking at the FBDs above, which of the following is the correct representation of the y-component of Newton's second law applied to
m1?
 Recall that
T1
 represents the magnitude of the tension force in the top cord and
Fg1
 represents the magnitude of the gravity force on
m1.
     Incorrect: Your answer is incorrect.

Incorrect. Both the tension in the bottom cord and the force of gravity act down in the negative y-direction.
Part 7 of 8
Apply Newton's second law (cont.).
The correct form for Newton's second law applied to
m1
 is
T1 − T2 − Fg1 = 0.
 Now let's find out whether or not any of the quantities in this expression are known. Recall from above that
T1
 is the unknown quantity we are to determine and that
T2 = m2g.

Is
Fg1
 a known quantity?
     Correct: Your answer is correct.
Correct. The quantity
Fg1
 is the magnitude of the upper block's weight. The magnitude of the weight of any object is given by
Fg = mg.
 Because we know both
m1
 and
g,
Fg1
 is known.
Substituting
T2 = m2g
 and
Fg1 = m1g
 into Newton's second law gives
T1 − T2 − Fg1 = T1 − m2g − m1g = 0,
and solving for
T1
 gives
T1 = m1g + m2g.
We have applied Newton's second law and substituted all the known quantities. The final step in the solution is to solve for the unknown quantity.
Part 8 of 8
Solve for the desired unknown quantity and substitute the numbers.
Our unknown quantity is
T1,
 the magnitude of the tension in the top cord. Solving for it gives
T1 = m1g + m2g
 = (m1 + m2)g
 = (5.30 kg + 2.70 kg)(9.80 m/s2)
 = 78.4 N.


Additional Problems

Question 4.2a:
The
5.30-kg
 block shown in the figure is suspended between two cords of negligible mass. The cord on the left is attached to a wall and is horizontal. A second cord attached to the block has tension
T = 72.0 N
 directed at an angle θ above the horizon. The block is in equilibrium.
Determine the tension
Tw
 in the cord attaching the block to the wall and the angle θ.
tension
Tw
 		= Correct: Your answer is correct. Correct. N
angle θ = Incorrect: Your answer is incorrect.
Incorrect. See Solution 4.2a. °

Question 4.2b:
The two masses shown in the figure are in equilibrium and the inclined plane is frictionless.
If
θ = 37.5°
 and
m2 = 3.43 kg,
 determine the mass of m1.
Incorrect: Your answer is incorrect.
Incorrect. See Solution 4.2b. kg

Question 4.2c:
The three masses shown in the figure are in equilibrium and the inclined plane is frictionless.
If
θ = 35.2°,
m2 = 2.00 kg,
 and
m3 = 4.60 kg,
 determine the mass
m1
 and the magnitude of the tension
T
 in the string connecting
m2
 and
m3.
mass
m1
 		= Incorrect: Your answer is incorrect.
Incorrect. See Solution 4.2c. kg
tension T = Incorrect: Your answer is incorrect.
Incorrect. See Solution 4.2c. N
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3. 0/15 points  |  Previous Answers was SerCP10 4.PLE.003. My Notes
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0/15
 
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

Prelecture Exploration: Contact Forces
Classmates Sarah and Kasim are study partners, and they are trying to understand forces between two objects in contact. There are many concepts to consider, including Newton's three laws, the masses of the two objects, and diagrams involving these contact forces.

In this simulation, they can observe the effects of an external force applied to two blocks in contact on a smooth surface. The masses of each block and the external force can be changed using the sliders. The direction of the external force can be set by clicking the "dir'n" button. Time and displacement are shown. After 10 seconds, the simulation stops. Note that the blocks might end up off the screen.

Display in a New Window
Part 1 of 14 - Contact Forces and Acceleration
Kasim asks Sarah to explain to her how two unattached but adjacent boxes accelerate when a contact force is exerted on one of them, and to help visualize the question, Kasim draws a diagram similar to this:
Suppose box 1 (m1) is cardboard and has oranges inside, whereas box 2 (m2) is a tin of breath lozenges. A finger applies a constant force of 0.1 N to the side of box 1. What should Sarah tell her friend Kasim about the acceleration of box 1 and box 2?
    


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4. –/6 points SerCP10 4.AE.007. My Notes
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EXAMPLE 4.7 Sled on a Frictionless Hill
(a) A sled tied to a tree on a friction less hill. (b) A diagram of forces acting on the sled.
Goal Use the second law and the normal force in an equilibrium problem.

Problem A sled is tied to a tree on a frictionless, snow-covered hill, as shown in figure (a). If the sled weighs 77.0 N, find the magnitude of the tension force T with arrow exerted by the rope on the sled and that of the normal force n with arrow exerted by the hill on the sled.

Strategy When an object is on a slope, it's convenient to use tilted coordinates, as in figure (b), so that the normal force n with arrow is in the y-direction and the tension force T with arrow is in the x-direction. In the absence of friction, the hill exerts no force on the sled in the x-direction. Because the sled is at rest, the conditions for equilibrium, ΣFx = 0 and ΣFy = 0 apply, giving two equations for the two unknowns—the tension and the normal force.
SOLUTION
Apply Newton's second law to the sled, with a with arrow = 0.
F with arrow = T with arrow + n with arrow + F with arrowg = 0
Extract the x-component from this equation to find T. The x-component of the normal force is zero, and the sled's weight is given by mg = 77.0 N.
Fx		 =T + 0 − mg sin θ = T − (77.0 N)(sin 30.0°) = 0
T = 38.5 N
Write the y-component of Newton's second law. The y-component of the tension is zero, so this equation will give the normal force.
Fy		 =0 + n − mg cos θ = n − (77.0 N)(cos 30.0°) = 0
n = 66.7 N
LEARN MORE
Remarks Unlike its value on a horizontal surface, n is less than the weight of the sled when the sled is on the slope. This is because only part of the force of gravity (the x-component) is acting to pull the sled down the slope. The y-component of the force of gravity balances the normal force.

Question Consider the same scenario on a hill with a steeper slope.
Would the magnitude of the tension in the rope get larger, smaller, or remain the same as before?
    

How would the normal force be affected?
    
PRACTICE IT
Use the worked example above to help you solve this problem. A sled is tied to a tree on a frictionless, snow-covered hill, as shown in Figure (a). If the sled weighs 76.0 N, find the magnitude of the tension force T with arrow exerted by the rope on the sled and that of the normal force n with arrow exerted by the hill on the sled.
T = N
n = N
EXERCISE HINTS:  GETTING STARTED  |  I'M STUCK!
Use the values from PRACTICE IT to help you work this exercise. Suppose a child of weight w climbs onto the sled. If the tension force is measured to be 58.0 N, find the weight of the child and the magnitude of the normal force acting on the sled.
w = N
n = N
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5. –/2 points SerCP10 4.CQ.013. My Notes
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–/2
 
In a tug-of-war between two athletes, each pulls on the rope with a force of 190 N. What is the tension in the rope?
N

If the rope doesn't move, what horizontal force does each athlete exert against the ground?
N
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6. –/4 points SerCP10 4.P.014. My Notes
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–/4
 
An object of mass m is dropped from the roof of a building of height h. While the object is falling, a wind blowing parallel to the face of the building exerts a horizontal constant force F on the object.
(a) How long does it take the object to strike the ground? Express the time t in terms of g and h.
t =


(b) Find an expression in terms of m and F for the acceleration ax of the object in the horizontal direction (taken as the positive x-direction).
ax =


(c) How far is the object displaced horizontally before hitting the ground? Answer in terms of m, g, F, and h.
Δx =


(d) Find the magnitude of the object's acceleration while it is falling, using the variables F, m, and g.
a =
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7. –/2 points SerCP10 4.P.030. My Notes
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A block of mass
m = 5.3 kg
 is pulled up a
θ = 22°
 incline as in the figure below with a force of magnitude
F = 36 N.
(a) Find the acceleration of the block if the incline is frictionless. (Give the magnitude of the acceleration.)
m/s2

(b) Find the acceleration of the block if the coefficient of kinetic friction between the block and incline is 0.11. (Give the magnitude of the acceleration.)
m/s2
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8. –/11 points SerCP10 4.P.029.MI.SA. My Notes
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–/11
 
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.

Tutorial Exercise
Assume the three blocks (m1 = 1.0 kg, m2 = 2.0 kg, and m3 = 3.3 kg) portrayed in the figure below move on a frictionless surface and a force F = 31 N acts as shown on the 3.3-kg block.
(a) Determine the acceleration given this system.

(b) Determine the tension in the cord connecting the 3.3-kg block and the 1.0-kg blocks.

(c) Determine the force exerted by the 1.0-kg block on the 2.0-kg block.

Step 1
(a) If we consider the system to consist of all three blocks plus the connecting rope, the total mass of the system is
mtotal = m1 + m2 + m3 = (No Response) seenKey

6.3

 kg.
The only horizontal external force acting on this system is the applied force directed toward the right, which is chosen as the positive x-direction. Thus, Newton's second law gives
ax = 
ΣFx
mtotal
 = 
(No Response) seenKey

31

 N
(No Response) seenKey

6.3

 kg
 
1 kg · m/s2
1 N
 = (No Response) seenKey

4.92

 m/s2.
Step 2
(b) In order to determine the tension in the rope, we define a new system consisting only of the block of mass m3. The external horizontal forces acting on this system are the applied force and the tension in the rope as shown in the free-body diagram below, where the normal force from the frictionless surface on the block of mass m3 is n with arrow3.
Let T be the tension in the rope. We have
ΣFx − T = m3ax
or, solving for the tension T,
T = ΣFx − m3ax
=
(No Response) seenKey

31

 N
 − 
(No Response) seenKey

3.3

 kg
(No Response) seenKey

4.92

 m/s2
= (No Response) seenKey

14.8

 N.
Step 3
(c) Now we consider a new system consisting only of m2, the 2.0-kg block. The horizontal force exerted on it by m1, the 1.0-kg block, is the external force F with arrow12 shown in the free-body diagram below.
In this diagram, n with arrow2 is the normal force of the frictionless surface on m2, the 2.0-kg block. Applying ΣFx = max from Newton's second law to this system gives
F12 = m2ax
=
(No Response) seenKey

2

 kg
(No Response) seenKey

4.92

 m/s2
 
 		= (No Response) seenKey

9.84

 N.
SUMMARY
From the total mass of the system and the force applied in the horizontal direction to the system, we calculated the system's acceleration using Newton's second law. We used this acceleration and two free-body diagrams for the blocks of mass m3 and m2 to determine (1) the tension in the rope connecting the first and third blocks and (2) the applied force exerted on the first block by the second block.
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