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Physics - College, section 1, Fall 2019
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Serway and Jewett - Principles of Physics 4/e (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 16 / 39

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

Last Saved : n/a Saving... ()

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Instructions

Principles of Physics: A Calculus-Based Text 4/e by Raymond A. Serway and John W. 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.

Question 1 is a traditional end-of-chapter symbolic question, which allows students to enter any mathematically equivalent answer. (For this demo, the solution is displayed immediately after the question has been submitted.)

Question 2 is an end-of-chapter question enhanced by adding conditional feedback to the numerical answers and a link to an Active Figure. The solution is display after the question has been answered.

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

Question 4, 5, and 6 are new Active Examples which help guide students through the process needed to master a concept.

Question 7 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 wherever the problem has randomized values.

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 points SerPOP4 5.P.022. My Notes

We will study the most important work of Nobel laureate Arthur Compton in a later chapter. Disturbed by speeding cars outside the physics building at Washington University in St. Louis, Compton designed a speed bump and had it installed. A car of mass m passes over a bump in a road that follows the arc of a circle of radius R, as in the figure below.

(a) What force does the road exert on the car as the car passes the highest point of the bump if the car travels at a speed v? (Use any variable or symbol stated above along with the following as necessary: g.)

(b) What is the maximum speed the car can have as it passes this highest point without losing contact with the road? (Use any variable or symbol stated above along with the following as necessary: g.)

v_max

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2. –/5 points was SerPOP4 4.P.032.AF. My Notes

Two objects with masses of 2.10 kg and 4.20 kg are connected by a light string that passes over a light frictionless pulley to form an Atwood machine as shown in Active Figure 4.12a.

Active Figure 4.12a

(a) Determine the tension in the string.
N

(b) Determine the acceleration of each object.

2.10 kg mass	m/s² upwards
4.20 kg mass	m/s² downwards

2.10 kg mass	m
4.20 kg mass	m

Refer to the Active Figure Simulation to review the concepts this question addresses.

Hint: Active Figure 4.12

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3. –/10 points was SerPOP4 4.AF.13. My Notes

Active Figure 4.13 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.

Start the simulation and observe the masses, as they accelerate to the right on a frictionless surface and eventually disappear off the screen.
Observe the on-screen values of time, distance and force and notice that the force remains constant throughout the motion.
Record the final values of distance and force at the finishing time of 10 seconds.

Exploration 2

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?
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.
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 vector F

to the right, and a contact force P₂₁ acting to the left. Newton's second law for mass m₁ gives:

F_x =

- P₂₁ = m₁a

(b) The applied force vector F

must make both blocks accelerate. Solve this expression for the magnitude of 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,

F_x = P₁₂ = m₂a.

(c) Solve this for the magnitude of the contact force acting on m₂:

|P₁₂| = N

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 magnitude of the contact force.

|P₁₂| = N

(e) Isolate m₁, apply Newton's second law, and solve for the magnitude of 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 4.AF.13

Suppose that the masses in the simulation are changed so that m₁ = 13.8 kg and m₂ = 6.9 kg. Suppose also that a new external force vector F

is applied so that the acceleration of the two blocks is 1.69 m/s².

(f) What is the magnitude of the applied force?

= 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 = N
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

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4. –/8 points SerPOP4 4.AE.01. My Notes

Example 4.1 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 Figure 4.4. The force vector F

₁ has a magnitude of 4.6 N, and vector F

₂ has a magnitude of 8.0 N. If

₁ = 20° and

₂ = 52°, 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 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 given by the following expression.

F_x = F_1x + F_2x = F₁ cos 20° + F₂ cos 52° = N

Next we find the component of the net force in the y direction.

F_y = F_1y + F_2y = -F₁ sin 20° + F₂ sin 52° = N

Now we use Newton's second law in component form to find the x and y components of acceleration.

a_x =

= m/s²

a_y = (Sigma F_y)/m

= m/s²

The acceleration has the following magnitude.

a =

= m/s²

Its direction is calculated relative to the positive x axis.

θ = tan^-1 $a_y / a_x$ = °

Exercise 4.1

Hints: Getting Started | I'm Stuck

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?

F_3x = N

F_3y = N

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5. –/7 points was SerPOP4 12.AE.01. My Notes

Example 12.1 A Block-Spring System

Problem A block with a mass of 280 g is connected to a light horizontal spring of force constant 5.1 N/m and is free to oscillate on a horizontal, frictionless surface.

If the block is displaced 7.4 cm from equilibrium and released from rest as in Active Figure 12.4, find the period of its motion.

Determine the maximum speed and maximum acceleration of the block.

Express the position, velocity, and acceleration of this object at t = 2 s, assuming that ϕ = 0.

Figure 12.4 (Interactive Example 12.1) A block-spring system that is released from rest at x_i = A. In this case, ϕ = 0, and therefore x = A cos ωt.

Strategy The situation (we assume an ideal spring) tells us to use the simple harmonic motion model.

Solution

Using Equation 12.13 we obtain the following equation.

T = 2pi sqrt(m/k) = 2pi sqrt((280 text( x 10)^(-3) text(kg))/(5.1 text( N/m))) =

We use Equations 12.17 and 12.18, with A = 7.4

10^-2 m.

v_max = ωA = (2pi)/TA =

m/s

a_max = ω²A = ((2pi)/T)^2A =

m/s²

From Equations 12.6, 12.15, and 12.16, we have the following.

x = Acos ωt = m

v = -ωAsin ωt = m/s

a = -ω²Acos ωt = m/s²

Exercise 12.1

Hints: Getting Started | I'm Stuck

What If? What if another mass of m = 140 g is placed on top of the 280 g mass? Although the 280 g mass is on a frictionless surface, there is friction between the mass m and the 280 g mass. The (static) friction is just enough to hold the two masses together if the acceleration is less than 1.58 m/s². What is the maximum oscillation amplitude for this system so that the top mass m will not fall off while it oscillates?

A = m

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6. –/5 points SerPOP4 20.AE.04. My Notes

Example 20.4 The Electric Potential of a Dipole

Problem An electric dipole consists of two charges of opposite sign but equal magnitude separated by a distance 2a as in Figure 20.9. The dipole is along the x axis and is centered at the origin.

Calculate the electric potential at any point P along the axis.

Calculate the electric field on the axis at points very far from the dipole.

Figure 20.9 An electric dipole located on the x axis.

Strategy Use Equations 20.12 and 20.16.

Solution

Using Equation 20.12, we have the following. (Use k_e for k_e, and q, a, and x as necessary.)

V=k_e sum_i q_i/r_i=k_e(q/(x-a)+(-q)/(x+a))=

Click here for help with symbolic formatting.

Using Equation 20.16 and the result from part A, we calculate the electric field at P.

E_x = -(partialV)/(partialx)=-d/(dx)((2k_eqa)/(x^2-a^2))=-2k_eqad/(dx)(x^2-a^2)^(-1)

E_x = (-2k_eqa)(-1)(x^2-a^2)^(-2)(2x)=(4k_eqax)/(x^2-a^2)^2

If P is far from the dipole so that x >> a, then a² can be ignored in the term x² - a² and E_x becomes the following. (Use k_e for k_e, and q, a, and x as necessary.)

(x >> a)

Comparing this result to that from one where the field is along a line perpendicular to the line connecting the charges, we see a factor of 2 difference between the results for the field far from the dipole. In such an example, we are looking at the field along a line perpendicular to the line connecting the charges. As we see in Figure 19.11, the vertical components of the field cancel because the point at which we evaluate the field is equidistant from both charges.

Therefore, only the very small horizontal components of the individual fields contribute to the total field. In this example, we are looking at the field along an extension of the line connecting the charges. For points along this line, the field vectors have components only along the line and the field vectors are in opposite directions. The point at which we evaluate the field, however, is necessarily closer to one charge than to the other. As a result, the field is larger than that along the perpendicular direction by a factor of 2.

Exercise 20.4

Hints: Getting Started | I'm Stuck

A charge +q is located at the origin. A charge -2q is at 4.55 m on the x axis.

(a) For what finite value(s) of x is the electric field zero?
m

(b) For what finite value(s) of x is the electric potential zero?

smaller value	m

larger value	m

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.

Listed below are some values that lead to specific error messages. The randomized length in the formula below is d. Try substituting the incorrect answer and see what feedback you get!

Part	Incorrect formula	Incorrect answer
(a)	-d*(1-sqrt(2))	1.88
(a)	d/3	1.52
(b)	-d*(1+sqrt(2))	-11.0
Part	Correct formula	Correct answer
(a)	-d*(1 + sqrt(2))	-11.0
(b1)	-d	-4.55
(b2)	d/3	1.52

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7. –/2 points SerPOP4 4.QQ.04. My Notes

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?

You are equally rich. You are. Your friend is.

Explain.

This answer has not been graded yet.

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