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WebAssign - College Phys Alt Version - 1/e (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 23 / 43

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

Last Saved : n/a Saving...  ()

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Question
Points
1 2 3 4 5 6 7 8 9 10 11 12
2/2 1/3 1/3 9/10 4/4 3/3 3/3 –/4 0/3 –/3 –/4 –/1
Total
23/43 (53.5%)

  • Instructions

    The College Physics Alternate Version question collection by WebAssign includes over 1700 questions covering every concept in the course, designed to work with any textbook (or no textbook at all). Questions in this collection feature feedback for every question part and detailed solutions for every question. Additionally, a series of multi-step tutorials are available to help strengthen problem-solving skills and conceptual understanding.

    Questions 2, 3, 7, and 10 below feature multi-step tutorials.

    Question 8 features physPad for symbolic answer entry. "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.

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. 2/2 points  |  Previous Answers WAColPhysAV1 2.P.002. My Notes
Question Part
Points
Submissions Used
1 2
1/1 1/1
1/100 1/100
Total
2/2
 
Find the following (in m) for the path in the figure below.
On a graph with a horizontal axis labeled x (m), a path begins at x = 2. The path then continues to the right to x = 10, then curves back to the left to x = 8, then curves to the right again, ending at an arrowhead at approximately x = 11.
(a)
the total distance traveled
Correct: Your answer is correct. seenKey

13

 m
(b)
the displacement from start to finish
Correct: Your answer is correct. seenKey

9

 m


Solution or Explanation
(a)
The distance is the total distance traveled regardless of direction.
8 m + 2 m + 3 m = 13 m
(b)
The displacement depends on the direction. Consider the right to be the positive direction and left to be negative. Then, add the displacements of each segment of the path.
8 m + (−2) m + 3 m = 9 m
Or, since displacement is the change in position, simply find the final position with respect to the origin of the axis minus the initial position.
11 m − 2 m = 9 m
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2. 1/3 points  |  Previous Answers WAColPhysAV1 2.P.017.Tutorial. My Notes
Question Part
Points
Submissions Used
1 2 3
0/1 0/1 1/1
1/100 1/100 1/100
Total
1/3
 
A remote controlled toy car starts from rest and begins to accelerate in a straight line. The figure below represents "snapshots" of the car's position at equal 0.5 s time intervals. (Assume the positive direction is to the right. Indicate the direction with the sign of your answer.)
(a)
What is the car's average velocity (in m/s) in the interval between t = 1.0 s to t = 1.5 s?
Incorrect: Your answer is incorrect. seenKey

1
Recall that the definition of average velocity is the change in position divided by the change in time. m/s
(b)
Using data from t = 1.0 s to t = 2.0 s, what is the car's acceleration (in m/s2) at t = 1.5 s?
Incorrect: Your answer is incorrect. seenKey

0.8
Recall that the definition of average acceleration is the change in velocity divided by the change in time. What is the initial velocity? What is the final velocity? Be careful about the change in time as well—if you measure velocities at "midpoints" within two separate intervals, what is the separation between those midpoints? m/s2
(c)
Is the car's speed increasing or decreasing with time?
     Correct: Your answer is correct.
Tutorial

Solution or Explanation
(a)
Apply the definition of average velocity, change in position divided by change in time, to the interval from 1.0 s to 1.5 s.
v = 
Δx
Δt
 = 
xf − xi
tf − ti
 = 
0.9 m − 0.4 m
1.5 s − 1.0 s
 = 1.0 m/s
(b)
Apply the definition of average acceleration, change in velocity divided by change in time, to the given interval. This requires an initial and final velocity. The initial velocity is the velocity in the interval from 1.0 s to 1.5 s, found in part (a). The final velocity in the interval from 1.5 s to 2.0 s can be found from:
v = 
Δx
Δt
 = 
xf − xi
tf − ti
 = 
1.6 m/s − 0.9 m/s
2.0 s − 1.5 s
 = 1.4 m/s.
The average acceleration is then given by
a = Δv/Δt.
 The time interval Δt is measured from the midpoint of the first interval to the midpoint of the second, and so is therefore 0.5 s.
a = 
Δv
Δt
 = 
vf − vi
Δt
 = 
1.4 m/s − 1.0 m/s
0.5 s
 = 0.8 m/s2
(c)
Since the velocity and acceleration are both in the same direction (the positive direction in this case), the car's speed (or magnitude of velocity) is increasing.
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3. 1/3 points  |  Previous Answers WAColPhysAV1 2.P.030.Tutorial. My Notes
Question Part
Points
Submissions Used
1 2 3
1/1 –/1 –/1
1/100 0/100 0/100
Total
1/3
 
You toss a tennis ball straight upward. At the moment it leaves your hand it is at a height of 1.5 m above the ground, and it is moving at a speed of 7.2 m/s.
(a)
How much time (in s) does it take for the tennis ball to reach its maximum height?
Correct: Your answer is correct. seenKey

0.735

 s
(b)
What is the maximum height (in m) above the ground that the tennis ball reaches?
m
(c)
When the tennis ball is at a height of 2.4 m above the ground, what is its speed (in m/s)?
m/s
Tutorial
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4. 9/10 points  |  Previous Answers WAColPhysAV1 2.P.044. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5 6 7 8 9 10
1/1 1/1 1/1 1/1 1/1 1/1 1/1 1/1 0/1 1/1
1/100 1/100 1/100 1/100 1/100 1/100 1/100 1/100 1/100 1/100
Total
9/10
 
An athlete is training on a 100 m long linear track. His motion is described by the graph of his position vs. time, below.
On a coordinate plane the horizontal axis is labeled t (s) and the vertical axis is labeled x (m). There are four connected line segments labeled A, B, C, and D. Segment A begins at (0, 0) and rises up and to the right passing through (5, 30) to reach (10, 60). Segment B begins at (10, 60) and continues horizontally to end at (30, 60). Segment C begins at (30, 60) and drops down and to the right to end at the point (45, 40). Segment D begins at the point (45, 40) and continues up and to the right passing through the points (50, 60) and (55, 80) to end at the point (60, 100).
(a)
For each segment of the graph, find the magnitude (in m/s) and direction of the athlete's velocity.
magnitude vA Correct: Your answer is correct. seenKey

6

 m/s direction vA Correct: Your answer is correct. seenKey

positive x

 magnitude vB Correct: Your answer is correct. seenKey

0

 m/s direction vB Correct: Your answer is correct. seenKey

The magnitude is zero.

 magnitude vC Correct: Your answer is correct. seenKey

1.33

 m/s direction vC Correct: Your answer is correct. seenKey

negative x

 magnitude vD Correct: Your answer is correct. seenKey

4

 m/s direction vD Correct: Your answer is correct. seenKey

positive x
(b)
What are the magnitude (in m/s) and direction of the athlete's average velocity over the entire 60 s interval?
magnitude Incorrect: Your answer is incorrect. seenKey

1.67
The initial time is 0 s and the final time is 60 s. Remember that average velocity is the change in position (from initial to final) divided by the change in time (from initial to final). Average velocity is not the same, necessarily, as instantaneous velocity. m/s direction Correct: Your answer is correct. seenKey

positive x


Solution or Explanation
(a)
Average velocity is defined as
v = Δx/Δt,
 the change in position over change in time. For each segment of the graph, calculate the change in x, the vertical axis, divided by the change in t, the horizontal axis. In other words, find the slope of each segment. The sign of the slope indicates the direction of the velocity.
vA = 
60 m − 0
10 s − 0
 = 6.0 m/s
vB = 
0 − 0
30 s − 10 s
 = 0.0 m/s
vC = 
40 m − 60 m
45 s − 30 s
 = −1.33 m/s
vD = 
100 m − 40 m
50 s − 35 s
 = 4.00 m/s
(b)
The overall average velocity is the overall change in position divided by the overall change in time.
v = 
100 m − 0
60 s − 0
 = 1.67 m/s
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5. 4/4 points  |  Previous Answers WAColPhysAV1 4.P.004. My Notes
Question Part
Points
Submissions Used
1 2 3 4
1/1 1/1 1/1 1/1
1/100 1/100 1/100 1/100
Total
4/4
 
The diagrams below show different objects of equal masses that are acted on by one or more forces. In the diagrams below, each force vector labeled
F
 has the same magnitude.
WebAssign Plot
WebAssign Plot
WebAssign Plot
WebAssign Plot
(a)
Which of the four objects shown has a net zero force acting on it?
    
(b)
Which object or objects have the largest magnitude of force? (Select all that apply.)

(c)
Which object or objects move with constant velocity? (Select all that apply.)

(d)
Which object or objects move with changing speed? (Select all that apply.)

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6. 3/3 points  |  Previous Answers WAColPhysAV1 4.P.019. My Notes
Question Part
Points
Submissions Used
1 2 3
1/1 1/1 1/1
2/100 1/100 2/100
Total
3/3
 
In each of the diagrams below, one or more external forces act on the indicated system. Which of the forces listed below are the reaction to the external force mentioned? (Select all that apply.)
(a)
the mower exerts a force
Correct: Your answer is correct.

(b)
the Moon exerts a force
Correct: Your answer is correct.

(c)
the negatively charged electron (labeled e in the diagram) exerts a force
Correct: Your answer is correct.



Solution or Explanation
(a)
The mower is the system. First we list all the forces outside of the system that exert a force on the system and identify the reaction force corresponding to each of the forces in the list.
  1. The handle exerts a force down and to the right on the mower. The reaction force is the mower exerting a force up and to the left on the handle.
  2. Earth exerts a downward gravitational pull on the mower. The reaction force is the mower exerting an upward pull on the Earth.
  3. The ground exerts an upward (normal) force on the mower. The reaction force is the mower exerting a downward force on the ground.
(b)
The Moon is the system. First we list all the forces outside of the system that exert a force on the system and identify the reaction force corresponding to each of the forces in the list.
  1. Earth exerts a force to the left on the Moon. The reaction force is the Moon pulling to the right on the Earth.
  2. The Sun exerts a force to the right on the Moon. The reaction force is the Moon exerting a force to the left on the Sun.
(c)
The electron is the system. First we list all the forces outside of the system that exert a force on the system and identify the reaction force corresponding to each of the forces in the list.
  1. The negative plate exerts a force to the right on the electron. The reaction force is the electron exerting a force to the left on the negative plate.
  2. Earth exerts a downward force on the electron. The reaction force is the electron exerting an upward force on the Earth.
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7. 3/3 points  |  Previous Answers WAColPhysAV1 4.P.033.Tutorial. My Notes
Question Part
Points
Submissions Used
1 2 3
1/1 1/1 1/1
1/100 1/100 1/100
Total
3/3
 
A 62-kg man stands on a bathroom scale inside an elevator. The scale measures in units of newtons.
(a)
The elevator accelerates upward from rest at a rate of 1.15 m/s2 for 1.50 s. What does the scale read during this 1.50 s interval?
Correct: Your answer is correct. seenKey

679

 N
(b)
The elevator continues upward at constant velocity for 8.50 s. What does the scale read now?
Correct: Your answer is correct. seenKey

608

 N
(c)
While still moving upward, the elevator's speed decreases at a rate of 0.800 m/s2 for 3.00 s. What is the scale reading during this time?
Correct: Your answer is correct. seenKey

558

 N
Tutorial

Solution or Explanation
(a)
When the man stands on the scale, his weight acts on the scale. The scale exerts an upward force on the man. It is this upward force that registers as the scale reading. At home, when you stand on the scale, the scale reading equals your weight as there are no other forces acting on you. However, when you and the scale are accelerating up or down, the net force acting on you changes.
Let us consider the free-body diagram for the man.
WebAssign Plot
In the free-body diagram,
mg,
 the weight of the man acts vertically down due to the earth's gravitational force. The scale exerts an upward force on him and this is represented by
Fscale
 in the free-body diagram. The direction of the acceleration is up as shown by the arrow labeled
a.
We take the up direction to be positive and write Newton's second law
Fnet, y = may
 for this situation and solve for
Fscale.
 Note that
ay
 is positive here as the elevator is accelerating upward.
Fscale − mg = may
Fscale = may + mg = (62 kg)(1.15 m/s2 + 9.8 m/s2) = 679 N
Note that the scale reads a higher value than the true weight of the man.
(b)
Now the elevator moves with constant velocity and the acceleration is zero. The free-body diagram is shown below.
WebAssign Plot
Now the right-hand side of the equation for Newton's second law is zero as the velocity is constant and we have
Fscale − mg = 0
Fscale = mg = (62 kg)(9.8 m/s2) = 608 N.
The scale reads the true weight of the man same as when he steps on the scale at home.
(c)
The elevator's speed now decreases, so the acceleration vector is opposite to the direction of motion. Since the acceleration is in the downward direction, the y-component of the acceleration vector is then
−0.800 m/s2.
Fscale − mg = −may
Fscale = −may + mg = (62 kg)(−0.800 m/s2 + 9.8 m/s2) = 558 N
WebAssign Plot
Note that the scale reads a lower value than the true weight of the man.
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8. –/4 points WAColPhysAV1 4.P.036. My Notes
Question Part
Points
Submissions Used
1 2 3 4
–/1 –/1 –/1 –/1
0/100 0/100 0/100 0/100
Total
–/4
 
The three diagrams below show a block of mass m being pulled or pushed at constant velocity along a table with a force
F.
 Assume the surfaces to be frictionless.
(a)
What is the magnitude of the normal force in each case? Use the following as necessary: g, F, and θ.
case (i) N=
case (ii) N=
case (iii) N=
(b)
How would your answer to part (a) change if, all else being the same, the object moved with constant acceleration?
    
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9. 0/3 points  |  Previous Answers WAColPhysAV1 4.P.051. My Notes
Question Part
Points
Submissions Used
1 2 3
0/1 0/1 –/1
1/100 1/100 0/100
Total
0/3
 
(a)
A block of mass m = 2.70 kg is suspended as shown in the diagram below.
Assume the pulley to be frictionless and the mass of the strings to be negligible. If the system is in equilibrium, what will be the reading of the spring scale in newtons?
Incorrect: Your answer is incorrect. seenKey

26.5
Did you draw a free-body diagram and identify all forces acting on the system? Did you apply Newton's Second Law to the system? N
(b)
Two blocks each of mass m = 2.70 kg are connected as shown in the diagram below.
Assume the pulley to be frictionless and the mass of the strings to be negligible. If the system is in equilibrium, what will be the reading of the spring scale in newtons?
Incorrect: Your answer is incorrect. seenKey

26.5
Did you draw a free-body diagram for each mass and identify all forces acting on the system? Did you apply Newton's Second Law to each mass? N
(c)
A block of mass m = 2.70 kg is in equilibrium on an incline plane of angle θ = 35.0° when connected as shown in the diagram below.
Assume the mass of the strings to be negligible. If the system is in equilibrium, what will be the reading of the spring scale in newtons?
N
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10. –/3 points WAColPhysAV1 6.P.050.Tutorial. My Notes
Question Part
Points
Submissions Used
1 2 3
–/1 –/1 –/1
0/100 0/100 0/100
Total
–/3
 
As shown in the figure below, a box of mass
m = 20.0 kg
 is released from rest (at position A) at the top of a 30.0° frictionless incline.
The box slides a distance
d = 3.60 m
 down the incline before it encounters (at position B) a spring and compresses it an amount
xC = 0.260 m
 (to point C) before coming momentarily to rest. Using energy content, determine the following.
(a)
the speed (in m/s) of the box at position B
vB = m/s
(b)
the spring constant (in N/m)
k = N/m
(c)
the physical quantity that is constant throughout the process
    
Tutorial
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11. –/4 points WAColPhysAV1 6.P.052. My Notes
Question Part
Points
Submissions Used
1 2 3 4
–/1 –/1 –/1 –/1
0/100 0/100 0/100 0/100
Total
–/4
 
A 1.1 kg mass starts from rest at point A and moves along the x axis subject to the potential energy shown in the figure below.
(a)
Determine the speed of the mass at points B, C, D.
point B m/s point C m/s point D m/s
(b)
Determine the turning points for the mass. (Select all that apply.)

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12. –/1 points WAColPhysAV1 6.P.076. My Notes
Question Part
Points
Submissions Used
1
–/1
0/100
Total
–/1
 
As shown in the figure below, a 2.25 kg block is released from rest on a ramp of height h.
When the block is released, it slides without friction to the bottom of the ramp, and then continues across a surface that is frictionless except for a rough patch of width 15.0 cm that has a coefficient of kinetic friction μk = 0.710. Find h (in m) such that the block's speed after crossing the rough patch is 3.90 m/s.
m
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