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Serway and Jewett - Physics for S&E, Mod Phys 10/e (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 48 / 81

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

Last Saved : n/a Saving...  ()

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  • Instructions

    Featuring optimized online homework in WebAssign, new context-rich and Think-Pair-Share problems, and sound educational pedagogy, the 10th edition of the market-leading Physics for Scientists and Engineers with Modern Physics, by Raymond A. Serway and John W. Jewett and published by Cengage Learning, adopts an integrative approach to course materials that seamlessly matches curated content to the learning environment for which it was intended—from in-class group problem solving to online homework that utilizes targeted feedback and tutorials.

    Questions 1 and 2 are Optimized Problems, which includes contextual randomizations and variability along with targeted feedback and in-depth solutions.

    Question 3 is an Interactive Video Vignette (IVV) question. IVVs encourage students to address their alternate conceptions outside of the classroom. They include online video analysis and interactive individual tutorials to address learning difficulties identified by PER (Physics Education Research).

    Question 4 is a Think-Pair-Share activity. Think-Pair-Share problems and activities are similar to context-rich problems, but tend to benefit more from group discussion because the solution is not straight forward as a single concept. Some Think-Pair-Share problems require the group to discuss and make decisions; other are more challenging by the fact that some of the information is not and cannot be known.

    Question 5 is an MCAT-Style Passage Problem Module, which are available only in WebAssign. These modules are modeled after the new MCAT exam's "passage problems." Each module starts with a text passage (often with accompanying photo/figures) followed by 5-6 multiple choice questions. The passage and the questions are usually not confined to a single chapter, and feedback is available with each question.

    Question 6 is a Life Science problem, which highlights the relevance of physics principles to those students taking the course who are majoring in one of the life sciences.

    Question 7 is a Context Rich problem. The Context Rich problems always address the student as "you" the individual in the problem, and have real-world connections (rather than discussing "books on planes" or "balls on strings"). They are structured like a short story and may not always explicitly identify the variable that needs to be evaluated.

    Question 8 is a PreLecture Exploration problem. PreLecture Explorations use HTML5 interactive simulations that allow students to make predictions, change parameters, and observe results. Each engaging simulation, based on a relevant scenario, is followed by conceptual and analytic questions, guiding students to a deeper understanding and helping promote a robust physical intuition. They are the perfect resource for flipped classrooms, for increasing student engagement and interest in the material prior to lecture, or for the first conceptual homework assignment.

    Question 9 has a Watch It link, which provides step-by-step instruction with short, engaging videos that are ideal for visual learners.

    Question 10 has a Master It tutorial to show students how to solve a similar problem in multiple steps by providing direction along with derivation.

    Question 11 is an Active Example, which guides students through the process needed to master a concept.

    Question 12 is an Analysis Model Tutorial, which guides students through every step of the problem-solving process, driving them to see the important link between the situation in the problem and the mathematical representation of the situation.

    Question 13 is a Conceptual Question designed to help students test their understanding of physical concepts as they work through each chapter.

    Question 14 is an Objective Question. These can be used as in-class polling questions or as pre-class reading questions to test students on foundational concepts.

    Also available to adopters is an additional collection of over 2,200 questions that feature feedback and tutorials, organized to match the table of contents of Serway and Jewett Physics for Scientists and Engineers, 10th edition. This collection can be added from "Free Additional Content" when creating your WebAssign course at no additional cost.

    Question 15 is from this exclusive collection. 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.

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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. 1/1 points  |  Previous Answers SerPSE10 5.5.OP.010. My Notes
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What is the difference (in N) in the weight of a 52.0 kg gazelle as measured in a location where
g1 = 9.8095
 meters per second squared, and a location where
g2 = 9.7808
 meters per second squared?
Correct: Your answer is correct. seenKey

1.49

 N


Solution or Explanation
The distinction between mass and weight was discovered after Jean Richer transported pendulum clocks from Paris, France, to Cayenne, French Guiana in 1671. He found that they quite systematically ran slower in Cayenne that in Paris. The effect was reversed when the clocks returned to Paris.
Let's use the definition of weight,
w = Fg = mg.
 Recall that mass does not depend on location, so the difference in weight is
ΔFg = m(g1 − g2).
Substituting values, we get
ΔFg = (52.0 kg)(9.8095 m/s2 − 9.7808 m/s2) = 1.49 N.
A precise balance scale, as in a doctor's office, reads the same in different locations because it compares you with the standard masses on its beams. A typical bathroom scale is not precise enough to reveal this difference.
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A vertical cable supports two vehicles, a 1,207 kg minivan and a 1,461 kg SUV, with the minivan directly above the SUV. The cable is attached to a crane, which lifts the vehicles with an acceleration 1.32 m/s2. At one moment the minivan is moving at a speed of 3.75 m/s. Assume the cable does not stretch.
(a)
How do the velocity and acceleration of the two vehicles compare?
     Correct: Your answer is correct.
(b)
What is the tension, Tlower, (in N) in the cable between the minivan and the SUV?
Correct: Your answer is correct. seenKey

16200

 N
(c)
What is the tension, Tupper, (in N) in the cable above the minivan?
Correct: Your answer is correct. seenKey

29700

 N


Solution or Explanation
(a)
Because the cable does not stretch, at any instant the two vehicles have the same (though changing) velocity and at all instants they have the same (constant) acceleration.
(b)
Write Newton's second law for the SUV, where Tlower is the tension in the lower cable.
Fy = ma → Tlower − mSUVg = mSUVa
Solve for the tension Tlower and substitute values.
Tlower = mSUV(a + g) = (1,461 kg)((1.32 + 9.80) m/s2) = 16,200 N
(c)
Consider both vehicles to be a single object with mass equal to the sum of the masses of the two vehicles. Write Newton's second law for this combined object, where Tupper is the tension in the upper cable.
Fy = ma → Tupper − (mminivan + mSUV)g = (mminivan + mSUV)a
Solve for the tension Tupper and substitute values.
Tupper = (mminivan + mSUV)(a + g) = ((1,207 kg) + (1,461 kg))((1.32 + 9.80) m/s2) = 29,700 N
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4/5
 
Watch the video vignette below, and be sure to follow any instructions when prompted during the videos. Then answer the questions below about the concepts discussed in the video.
(a)
As you saw in the video, the two bullets in the experiment were fired from the same rifle. Assume the bullets also had the same mass and were fired from identically prepared cartridges. From this, what can we conclude about the initial momentum of the bullets just before they struck the blocks?
     Incorrect: Your answer is incorrect.
(b)
As you saw in the video, the two blocks had about the same mass, and both were initially at rest. Which of the following is true about the two bullet–block systems, just after the bullets collided with their respective blocks?
    
(c)
Which of the following statements provides correct reasoning for why both bullet–block systems reached the same maximum height?
     Correct: Your answer is correct.
(d)
Which of the following statements are true about the energy of the two bullet–block systems immediately after the collisions? (Select all that apply.)
Correct: Your answer is correct.

(e)
Imagine comparing the two blocks after the experiment. In which block would we expect the bullet to be embedded the greatest distance?
     Correct: Your answer is correct.
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Activity A simple procedure can be followed to measure the coefficient of friction using the technique discussed in Example Experimental Determination of μs and
μk.
 Lay your book down on a table and place a coin on the cover far from the spine. Slowly open the book cover so that it forms an inclined plane down which the coin will slide. Watch carefully and stop opening the cover at the instant the coin begins to slide.
(a)
Measure this critical angle that the book cover makes with the horizontal with a protractor. From this angle, determine the coefficient of static friction between the coin and the cover.

Key: The coefficient of static friction is given in the Example Experimental Determination of μs and μk, using the equation
μs = tan(θc),
 where
θc
 is the critical angle. Answers will vary, depending on the coin and the condition of the book cover.

Score: 1 out of 1

Comment:

(b)
Place a loop of tape between two coins and repeat the procedure above for the two-coin stack. How does the coefficient of static friction for the stack compare to that for the single coin?
     Correct: Your answer is correct.


Solution or Explanation
(a)
The coefficient of static friction is given in the Example Experimental Determination of μs and μk, using the equation
μs = tan(θc),
 where
θc
 is the critical angle. Answers will vary, depending on the coin and the condition of the book cover.
(b)
Ideally, the coefficient should not depend on the number of coins.
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5/6
 
The larvae of several species of antlion are known to dig cone-shaped holes in sand to trap their prey. The pits, typically 5.00–8.00 cm across, are excavated by the larva until the slopes are at a "critical angle of repose," or the angle at which the sand will begin to slide at the slightest disturbance, which is 34.0° for sand. This angle is obtained by computing the arctangent of the coefficient of static friction of the sand. When an ant or other insect begins to climb down the hole, it quickly loses its footing, at the same time causing the slope to become unstable. The antlion larva uses its torso to fling sand in the direction of the prey to ensure that it continues to slide toward the bottom of the pit, where it is consumed by the antlion larva.
(a) antlion
(b) antlion sandpit, with antlion larva hiding at its center
(c) cross-section of an antlion pit

  • (a)

    Figure (c) shows a cross-sectional view of a cone-shaped antlion pit that is 8.00 cm in diameter. The volume of a cone is given by
    V = πr2
    h
    3
    ,
     where r is the radius of the circular base of the cone and h is its height. If the density of sand is 1,280 kg/m3, what is the mass of sand (in g) that the antlion needs to excavate to dig this hole?
         Correct: Your answer is correct.

  • (b)

    The antlion digs its hole by first marking a circular area, then carrying or flicking sand from the deepening hole to just beyond the marked area. The center of mass of a cone of height h is located a distance
    h
    4
     from its base, and the work done by the antlion in removing the sand as it is excavating the pit is equal to the work done to raise the center of mass of the cone-shaped volume of sand to the undisturbed surface of the sand just beyond the hole. If an antlion has excavated 35.0 g of sand from a conical hole, what is the work done by the antlion (in mJ)? (Assume that excavating the sand in this instance results in a conical hole of diameter 6.76 cm, sloped at 34.0° as in figure (c).)
         Incorrect: Your answer is incorrect.

    How is the work done calculated in this case? How does the geometry of the hole affect your answer?

  • (c)

    As the antlion excavates the hole, it can sometimes dig too deep, creating a slope at an angle with the horizontal that is larger than the critical angle of repose. At this point, the sand slides down the slope until the critical angle is once again restored. If the antlion has dug a hole that momentarily has an angle of 40.5° with the horizontal, what is the magnitude of the net force (in N) on a grain of sand with a mass of 0.630 mg at the moment just before it begins to slide?
         Correct: Your answer is correct.

  • (d)

    The antlion sees an ant at the edge of the hole, and attempts to dislodge it by flicking sand from the bottom of the pit, at the apex of the inverted cone, toward the ant. If some of the grains of sand are launched at an angle of 45.0° from the horizontal, with what speed (in m/s) must the antlion launch the grains of sand to hit the ant? (Assume the hole has a diameter of 8.00 cm, as in figure (c).)
         Correct: Your answer is correct.

  • (e)

    If the antlion is able to either increase the speed with which it flicks the grains of sand by 10%, or vary the angle of launch by 10% from its initial 45.0° launch angle, which will result in a longer range for the flicked sand?
         Correct: Your answer is correct.

  • (f)

    Some of the grains of sand are launched directly upward by the antlion. What is the mathematical expression for the maximum height above the launch point of these sand grains?
         Correct: Your answer is correct.


Solution or Explanation
Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values.
(a)
The mass of an object is given by
m = ρV,
 where
ρ
 is the object's density and V is its volume. From the formula given in the problem statement, the mass of sand filling a conical volume is
m = ρV = ρ
πr2h
3
.
Now, examining the right triangle formed by the height and radius of the hole, we see that
tan(θ) = 
h
r
,
 or
h = r tan(θ).
 The mass of the sand in the conical hole is therefore
m = ρV = ρ
πr2
3
(r tan(θ)) = 
ρπr3 tan(θ)
3
.
Substituting numerical values, we obtain
m = 
(1,280 kg/m3)π(0.040 m)3 tan(34.0°)
3
 = 0.0579 kg = 57.9 g.
(b)
The work done by the antlion is equal to the change in potential energy of the center of mass of the conical volume of sand, or
W = ΔUg = mgΔhCM.
The center of mass of the inverted cone is located a distance
h
4
 below the surface of the sand, so
ΔhCM = 
h
4
 = 
r tan(θ)
4
.
The work done is then
W = mgΔhCM = 
mgr tan(θ)
4
 
 = 
(0.0350 kg)(9.80 m/s2)(0.0338 m) tan(34.0°)
4
 
 = 1.95 ✕ 10−3 J = 1.95 mJ.
(c)
The figure below shows a free-body diagram for a grain of sand on the slope. Selecting the coordinate system shown in the figure, we write the components of Newton's second law as
Fx		 = mg sin(θ) − fs = ma
 
Fy		 = n − mg cos(θ) = 0.
θ = 40.5°
The net force in the y-direction is zero, but there is a net force in the x-direction. From the second equation, we have
n = mg cos(θ)
 and, because the grain of sand is just ready to slide down the hill,
fs, max = μsn.
 Therefore, we have
Fnet = mg sin(θ) − fs, max = mg sin(θ) − μsmg cos(θ)
= mg(sin(θ) − μs cos(θ)).
From the problem statement,
μs = tan(34.0°) = 0.6745.
Substituting this and the given numerical values into the net force equation gives
Fnet = (0.630 ✕ 10−6 kg)(9.80 m/s2)[sin(40.5°) − (0.6745)cos(40.5°)]
 = 8.43 ✕ 10−7 N.
(d)
The figure below shows the relationship between the launch point of the sand and the position of the ant (circle at the top of the triangle). The equations of motion for the projectile are, with
xi = yi = 0,
x = (v cos(θi))t
y = (v sin(θi))t − 
1
2
gt2.
Solving the first equation for the time t gives
t = 
x
v cos(θi)
.
Substituting this expression into the y-equation then gives
y = (v sin(θi))
x
v cos(θi)
 − 
1
2
g
x
v cos(θi)
2
 
 = x tan(θi) − 
gx2
2v2 cos2(θi)
.
Now, from the figure above, we have
y = x tan(34.0°),
 so the above equation becomes
x tan(34.0°) = x tan(θi) − 
gx2
2v2 cos2(θi)
 
x(tan(θi) − tan(34.0°)) = 
gx2
2v2 cos2(θi)
Solving for v then gives
v = 
gx
2 cos2(θi)(tan(θi) − tan(34.0°))
 
 = 
(9.80 m/s2)(0.04 m)
2(cos(45.0°))2(tan(45.0°) − tan(34.0°))
 
 = 1.10 m/s.
(e)
If the initial and final y-coordinates of a projectile are equal, then the range of the projectile is given by
R = 
v2 sin(2θ)
g
.
In the case of the grain of sand launched by the antlion, the projectile is landing on a horizontal surface at a height h above its initial launch point, so the expression for range will be more complicated. However, the above equation illustrates the point that the maximum range of a projectile is obtained using a launch angle of 45.0°. Therefore, any variation from this launch angle will result in a shorter range for the projectile, whereas an increase in its launch speed will lead to an increase in its range.
(f)
The maximum height of a projectile can be obtained either from the kinematic equations or from conservation of mechanical energy in the projectile–Earth system. We use the latter approach, setting the reference configuration of gravitational potential energy as when the grain of sand is at the launch point. Then, since the projectile momentarily comes to rest at maximum height, conservation of energy gives
ΔK + ΔU = 0
 
0 − 
1
2
mv2
 + (mgh − 0)		 = 0.
Solving for h then yields
h = 
v2
2g
.
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Arguably the most hazardous portion of a parachute jump is landing on solid ground, when the skydiver is abruptly brought to a stop. Among injuries that can result during a hard landing are a dislocation of the ankle or the knee, or a fracture of the twin lower leg bones, the tibia and fibula, which can withstand a maximum force of 1.20 kN. One method of minimizing the impact of the landing is for the skydiver to bend his knees to extend the time of impact and reduce the average force experienced by the leg bones.
(a)
Suppose a skydiver's parachute does not operate properly and the downward speed of the diver as he lands is 32 kilometers per hour. What is the minimum distance (in m) through which the 94.0 kg skydiver must decelerate by bending his knees during impact to avoid fracturing his leg bones?
Correct: Your answer is correct. seenKey

3.09

 m
(b)
Based on your answer to part (a), can the skydiver safely land by only bending his knees?
     Correct: Your answer is correct.


Solution or Explanation
Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values.
(a)
From Newton's second law, the maximum acceleration the tibia and fibula can withstand during landing is given by
a = 
F
m
 = 
1.20 ✕ 103 N
94.0 kg
 = 12.8 m/s2
We can obtain the minimum distance required by solving
vxf2 = vxi2 + 2ax(xf − xi)
 for motion with constant acceleration,
vyf2 = vyi2 + 2aΔy,
 for
Δy,
 with 32 km/h = 8.89 m/s.
|Δy| = 
vyf2 − vyi2
2a
 = 
0 − (8.89 m/s)2
2(12.8 m/s2)
 = 3.09 m
(b)
No, he cannot land safely just by bending his knees. The length required is longer than the length of the human body. In such a situation, skydivers initiate a "parachute landing fall" in which they throw their bodies sideways to distribute the force of impact during landing.
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You are working for a rocket-launching company that is preparing to launch human space travelers into low-Earth orbit. In test flights, the company is having difficulty with a particular piece of electronic equipment, which is repeatedly damaged by the vibrations of the spacecraft as it launches. In previous launches, the equipment has been kept in a storage locker on the spacecraft, and apparently the vibrations from the walls of the locker have damaged the equipment. Your director has an idea for which he would like you to submit a design. He wants to hang the equipment on a single wire from the forward bulkhead of the spacecraft, surrounded by soft foam. When the spacecraft sits on the launcher nose-up, the equipment will hang downward from its attachment point and maintain this orientation as the spacecraft accelerates straight upward. The equipment and wrapping material have a mass of 18.5 kg. Beginning from rest, the rocket will reach a height of 45.0 km from the surface of the Earth in a time interval of 30.0 s. You need to determine the tension that the wire must be able to withstand in order that it not break during launch. (Give the tension in N.)
Correct: Your answer is correct. seenKey

2030

 N


Solution or Explanation
Conceptualize Consider the example in which we weighed a fish in an elevator, where the fish was hanging from a spring scale that was attached to the ceiling of an accelerating elevator. This is the same problem! The fish becomes the equipment package and the elevator accelerating upward becomes the spacecraft.
Categorize The equipment package is modeled as a particle under a net force. There is no information in the problem about time variation of the acceleration of the spacecraft during its launch, so let's assume that it is constant and then we will remark on other possibilities in the Finalize step. Therefore, we will use the particle under constant acceleration model for the package.
Analyze The forces on the equipment package are that due to the tension in the string and that due to the gravitational force, which are in opposite directions as the spacecraft rises vertically. Therefore, from Newton's second law, the tension in the wire is found as follows.
(1)    
Fy
 = may → T − Fg = may → T = Fg + may = mg + may = m(g + ay)
From
rf = ri + vit + 
1
2
at2
 in the particle under constant acceleration model,
(2)    yf = yi + vyit + 
1
2
ayt2 → ay = 
2(yf − yi − vyit)
t2
.
Substitute for the initial conditions.
(3)    ay = 
2(yf − 0 − 0)
t2
 = 
2yf
t2
Substitute equation (3) into equation (1).
T = m 
g + 
2yf
t2
Substitute numerical values.
T = (18.5 kg) 
9.80 m/s2 + 
2(4.50 ✕ 104 m)
(30.0 s)2
 = 2.03 ✕ 103 N
Finalize This is the minimum tension that the wire must withstand if it is not to break as the spacecraft accelerates. We do not know if the acceleration is truly constant; there may be times when it is larger than that suggested by equation (3). Therefore, to give a margin of error, the breaking strength of the wire should be significantly larger than this value.]
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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: Atwood Machine
When two massive objects are suspended on opposite sides of a pulley, the arrangement is called an Atwood machine. Two friends, Makena and Dani, are studying physics together, and they decide to delve further into understanding this arrangement.

In this simulation, the cord that connects the two masses is ideal; that is, it is inextensible (does not stretch) and its mass is negligible. Also, the pulley is frictionless with negligible mass, and any air resistance can be neglected. The forces and accelerations of the two masses are shown, along with the tension in the cord.

The sign convention for motion is defined as shown by the arrows. Makena and Dani can change the mass of m1 and m2 by using the sliders. They then can click "start" to set the system in motion.


Click here to open the simulation in a new window.
Part 1 of 13 - Accelerations in an Atwood Machine
Dani and Makena examine the figure shown.
Which of Dani's statements is correct, given what she knows about this system?
     Correct: Your answer is correct.
Dani is correct. She realizes that the rope connects the masses, so if one accelerates, both will accelerate, but she has no knowledge of the masses involved, so nothing definite can be said about the motion of the system without knowing the masses.
Part 2 of 13 - Accelerations in an Atwood Machine
Now the friends consider the situation for which the mass m1 is initially moving downward. Makena and Dani discuss the acceleration in this case. Which of Makena's statements could be correct?
     Correct: Your answer is correct.
Makena is correct. She realizes that the rope connects the masses, so if one accelerates, both will accelerate, but she has no knowledge of the masses involved, so nothing definite can be said about the motion of the system without knowing the masses.
Part 3 of 13 - Acceleration of the System
Makena and Dani finally realize that because the masses are connected by an inextensible cord, if one goes down, the other must be going up.

The two friends decide to use a simulation to gain a better understanding of an Atwood machine. Dani sets the masses equal to any particular values, with m2 > m1. What can Makena say about the magnitudes of the acceleration of m1 and m2?
     Correct: Your answer is correct.
Makena is correct. The accelerations are equal for any values of m1 and m2. What is the reason they are the same?
Part 4 of 13 - Acceleration of the System
The friends now know that the acceleration magnitudes are the same. Which of Dani's statements explains the reason the acceleration of m1 is equal to the acceleration of m2 for any nonzero values of m1 and m2?
     Correct: Your answer is correct.
Dani is correct. Because the two masses are connected by the cord, they must move together; otherwise, the cord would break. Therefore, the magnitudes of their displacements, velocities, and accelerations must be the same.
Part 5 of 13 - Tension in the Cord
For m2 > m1, Makena and Dani take T1 and T2 to be the magnitudes of the tension in the cord connected to m1 and m2, respectively. Which of Makena's statements is true?
     Correct: Your answer is correct.
Makena is correct. For a frictionless pulley and cord of negligible mass, the tension is the same throughout the cord, so the magnitudes of the tension forces acting on the two masses are the same.
Part 6 of 13 - Tension in the Cord
Dani and Makena consider the case for which m2 > m1. If the tension is constant throughout the cord, how can Dani explain why it is that the two masses accelerate?
     Correct: Your answer is correct.
Dani is correct. For example, for m1, the net force is the vector sum of the weight of m1 and the tension in the cord connected to it, which, for
m1 ≠ m2
 and
m1 ≠ 0,
 is nonzero, so there is a resulting acceleration.
Part 7 of 13 - Limiting Behavior
Dani and Makena now consider the limiting case for which m1 = 0, whereas m2 is some constant value. Which of Makena's statements is false?
     Incorrect: Your answer is incorrect.

This statement is incorrect because Makena is correct, and you are looking for an incorrect response. Because there is no tension force, the only force acting on the system is the gravitational force on m2, and the magnitude of the gravitational acceleration is g.
Part 8 of 13 - Limiting Behavior
The women now consider the limiting case where
m2 → ∞
, whereas m1 remains some constant (fixed) value. What value does Dani say the magnitude of the acceleration approaches?
    
Part 9 of 13 - Mathematical Expressions
Dani asks Makena to write the expression for Newton's second law for m1, using the sign convention as used in the simulation. Which expression is correct?
    
Part 10 of 13 - Mathematical Expressions
Makena now challenges Dani to write an expression for Newton's second law for m2, using the sign convention as used in the simulation. Which expression is correct?
    
Part 11 of 13 - Mathematical Expressions
Makena asks Dani how to use Newton's second law equations for m1 and m2,
ΣF = T − m1g = m1a
 and
ΣF = −T + m2g = m2a,
 respectively, to determine the magnitude of the acceleration of system. Which is the correct response?
    
Part 12 of 13 - Mathematical Expressions
How can Makena use the expression for a,
a = 
m2 − m1
m1 + m2
g
, and Newton's second law to find the tension in the cord?
    
Part 13 of 13 - Analyze
The friends now feel ready to try a problem.

Suppose an Atwood machine has a mass of m1 = 6.5 kg and another mass of m2 = 9.5 kg hanging on opposite sides of the pulley. Assume the pulley is massless and frictionless, and the cord is ideal. Determine the magnitude of the acceleration of the two objects and the tension in the cord.
a = (No Response) seenKey

1.84

 m/s2
T = (No Response) seenKey

75.6

 N
Going Further: Measuring g
An Atwood machine can be used to measure the acceleration of gravity, g. For m2 > m1, the acceleration of the two masses is given by
a = g
m2 − m1
m2 + m1
.
 In a particular experiment, different values of m1 and m2 are tested, and the experimental value of a is found by measuring the time it takes for m2 to drop a fixed distance, d, then using
d = 
1
2
at2.
 The experimental value of g can be found using
a = g
m2 − m1
m2 + m1
 and the corresponding values of m1, m2, and a.

Imagine that this experiment has been run for several different values of m1 and m2. Which of the following is a reasonable method for Makena to use to determine the experimental value of g? (Select all that apply.)

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A force F with arrow applied to an object of mass m1 produces an acceleration of 3.60 m/s2. The same force applied to a second object of mass m2 produces an acceleration of 1.30 m/s2.
(a) What is the value of the ratio m1/m2?
Incorrect: Your answer is incorrect. seenKey

0.361
Newton's second law relates all of the given quantities. Write this equation once with the 1 subscript attached to the symbols and again with the 2 subscript. Then do some algebra to relate m1/m2 to the accelerations.

(b) If m1 and m2 are combined into one object, find its acceleration under the action of the force F with arrow.
Correct: Your answer is correct. seenKey

0.955

 m/s2
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An electron of mass 9.11 10-31 kg has an initial speed of 4.00 105 m/s. It travels in a straight line, and its speed increases to 7.80 105 m/s in a distance of 4.40 cm. Assume its acceleration is constant.
(a) Determine the magnitude of the force exerted on the electron.
N

(b) Compare this force (F) with the weight of the electron (Fg), which we ignored.
F
Fg
 = Correct: Your answer is correct. seenKey

5.20e+11

Need Help? Master It

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15/21
 

One Block Pushes Another

Two blocks of masses m1 and m2, with m1 > m2, are placed in contact with each other on a frictionless, horizontal surface as in figure (a). A constant horizontal force
F
 is applied to m1 as shown.
(a) A force is applied to a block of mass m1, which pushes on the second block of mass m2.
(b) The forces acting on m1.
(c) The forces acting on m2.
(a)
Find the magnitude of the acceleration of the system.
SOLUTION
Conceptualize Conceptualize the situation by using figure (a) and realize that both blocks must experience because they are in contact with each other and remain in contact throughout the motion.
Categorize We categorize this problem as one involving a particle under a net force because a force is applied to a system of blocks and we are looking for the of the system.
Analyze First model the combination of two blocks as a single particle under a net force. Apply Newton's second law to the combination in the x-direction to find the acceleration. (Use the following as necessary: F, m1, and m2.)
Fx = F = (m1 + m2)ax
(1)    ax =
ab​
Finalize The acceleration given by Equation (1) is the same as that of a single object of which following mass, subject to the same force?
    
(b)
Determine the magnitude of the contact force between the two blocks.
SOLUTION
Conceptualize The contact force is internal to the system of two blocks. Therefore, we find this force by modeling the whole system (the two blocks) as a single particle.
Categorize Now consider each of the two blocks individually by categorizing each as a particle under a net .
Analyze
We construct a diagram of forces acting on the object for each block as shown in figures (b) and (c), where the contact force is denoted by
P.
 From figure (c), we see that the only horizontal force acting on m2 is the contact force
P12,
 which is directed to the right and is exerted by which mass?
    
Apply Newton's second law to m2. (Use the following as necessary: ax, g, and m2.)
(2)    
Fx = P12 =
Substitute the value of the acceleration ax given by Equation (1) into Equation (2). (Use the following as necessary: F, m1, and m2.)
(3)    P12 = m2ax =
Finalize This result shows that the contact force P12 is less than the applied force F. The force required to accelerate block 2 alone must be the force required to produce the same acceleration for the two-block system.
To finalize further, let us check this expression for P12 by considering the forces acting on m1, shown in figure (b). The horizontal forces acting on m1 are the applied force
F
 to the and the contact force
P21
 to the (the force exerted by m2 on m1). From Newton's third law,
P21
 is the reaction force to
P12,
 so P21 = P12.
Apply Newton's second law to m1. (Use the following as necessary: ax, g, and m1.)
(4)    
Fx = F − P21 = F − P12 =
Solve for P12 and substitute the value of ax from Equation (1). (Use the following as necessary: F, m1, and m2.)
P12 = F − m1ax = F − m1
F
m1 + m2
 =
This result with Equation (3), as it must.
EXERCISE
Three blocks are in contact with one another on a frictionless, horizontal surface as in the figure below. A horizontal force
F
 is applied to m1, where m1 = 1.70 kg, m2 = 3.73 kg, m3 = 4.92 kg, and F = 19.0 N. (Take the +x-direction to be to the right. Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)
Figure from book
(a)
Find the acceleration of the blocks. (Enter your answer in m/s2. Indicate the direction with the sign of your answer.)
m/s2
(b)
Find the net force on each block. (Enter your answers in N. Indicate the direction with the signs of your answers.)
F1, net = N F2, net = N F3, net = N
(c)
Find the magnitudes of the contact forces between the blocks (in N).
P12 = N P23 = N
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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.

Analysis Model Tutorial
A bag of cement weighing 292 N hangs in equilibrium from three wires as suggested in the figure. Two of the wires make angles
θ1 = 61.0°
 and
θ2 = 41.0°
 with the horizontal. Assuming the system is in equilibrium, find the tensions T1, T2, and T3 in the wires.
Part 1 of 8 - Conceptualize:
Look carefully at the figure. Which force do you expect to be larger,
T1 or T2?
 Should we expect that
T1 and T2
 will each be larger than the weight of the bag because only a component of their magnitude is in the vertical direction? Or should they each be smaller because they work together to hold up the bag? Or could one force be larger than the weight and one smaller? We'll choose a standard coordinate system with x horizontal and y vertical.
Part 2 of 8 - Categorize:
(1) What analysis model correctly describes the knot, that is, the point at which all three forces meet, and is most useful for solving this problem?
    
Part 3 of 8
(2) The ring at the top of the strap, where the rope carrying force
T3
 is attached, is also a particle in equilibrium. Therefore, the tension
T3 = 292 Correct: Your answer is correct. seenKey

292

 N.
Part 4 of 8 - Analyze:
(3) Based on the analysis model in question (1), which equation below correctly describes the relationship between the x components of the forces on the knot?
     Correct: Your answer is correct.
Correct. This is the correct expression relating the horizontal components of the forces.
(4) Based on the analysis model in question (1), which equation below correctly describes the relationship between the y components of the forces on the knot?
     Correct: Your answer is correct.
Correct. This is the correct expression relating the vertical components of the forces.
Part 5 of 8 - Analyze: (cont.)
(5) Based on the correct choice in question (3), which equation below correctly describes the relationship between
T2 and T1?
     Correct: Your answer is correct.
Correct. This is the correct rearrangement of the equation in question (3).
Part 6 of 8 - Analyze: (cont.)
(6) Combine the equations in the correct choices from questions (4) and (5) to eliminate
T2.
 Which equation below is the correct expression of
T1
 in terms of
T3?
     Correct: Your answer is correct.
Correct. This is the correct combination of the two equations.
Part 7 of 8 - Analyze: (cont.)
(7) Based on the correct choice in question (6) above, substitute numerical values to find the value of
T1.
T1 = 
T3
sin θ1 + cos θ1 tan θ2
 
 = 
Incorrect with feedback: Your answer is incorrect. Click to see feedback.

This feedback is based on your last submitted answer, not on your current answer.

What value of T3 did we find earlier?
seenKey

292

 N
sin Incorrect with feedback: Your answer is incorrect. Click to see feedback.

This feedback is based on your last submitted answer, not on your current answer.

What angle does the wire with force T1 make with the ceiling?
seenKey

61.0

 ° + 
cos Incorrect with feedback: Your answer is incorrect. Click to see feedback.

This feedback is based on your last submitted answer, not on your current answer.

What angle does the wire with force T1 make with the ceiling?
seenKey

61.0

 °
tan Incorrect with feedback: Your answer is incorrect. Click to see feedback.

This feedback is based on your last submitted answer, not on your current answer.

What angle does the wire with force T2 make with the ceiling?
seenKey

41.0

 °
 
 =  Incorrect with feedback: Your answer is incorrect. Click to see feedback.

This feedback is based on your last submitted answer, not on your current answer.

Check to see if the individual values you entered are correct and then check your numerical calculation.
seenKey

225

 N


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A person holds a ball in her hand.
(a) Identify all the external forces acting on the ball and the Newton's third-law reaction force to each one. (Select all that apply.)
Incorrect: Your answer is incorrect.

(b) If the ball is dropped, what force is exerted on it while it is falling? Identify the reaction force in this case. (Ignore air resistance.)
    
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In the figure below, a locomotive has broken through the wall of a train station. During the collision, what can be said about the force exerted by the locomotive on the wall?
     Incorrect: Your answer is incorrect.
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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 = 8.7 N. Here, A has a mass mA = 10.6 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 0.04. (Assume
FP
 acts in the +x direction.)
(a) What is the magnitude of the acceleration of the two boxes?
Correct: Your answer is correct. seenKey

0.102

 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
Tutorial
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