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WebAssign - College Physics 1/e (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 13 / 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
1/2 2/3 –/3 7/10 1/4 1/3 –/3 –/4 1/3 –/3 –/4 0/1
Total
13/43 (30.2%)

  • Instructions

    College Physics is our new premiere collection of algebra-based physics content, available to be used with any textbook (or no textbook at all!).

    Available questions in WebAssign have been authored and peer-reviewed by experienced physics educators, and have been written specifically for an online interactive environment. Each question includes answer feedback developed specifically to address common student misconceptions and a complete detailed solution. A series of multi-step tutorials guide students through the problem-solving process and are offered as question hints or assignable as individual items. This unique feature allows instructors the flexibility to adapt their assignments to best help their students learn. 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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1. 1/2 points  |  Previous Answers WebAssignAlgPhys1 2.1.002. My Notes
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0/1 1/1
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1/2
 
Find the following for the path in the figure below.
(a) the total distance traveled
Incorrect: Your answer is incorrect. seenKey

13
The question asks for the total distance traveled, regardless of direction. 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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2/3
 
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 the interval between t = 0.5 s to t = 1.0 s?
Correct: Your answer is correct. seenKey

0.6

 m/s

(b) Using data from t = 0.5 s to t = 1.5 s, what is the car's acceleration at t = 1.0 s?
m/s2

(c) Is the car's speed increasing or decreasing with time?
     Correct: Your answer is correct.

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3. –/3 points WebAssignAlgPhys1 2.7.001.Tutorial. My Notes
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–/1 –/1 –/1
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–/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 8.0 m/s.
(a) How much time does it take for the tennis ball to reach its maximum height?
s

(b) What is the maximum height above the ground that the tennis ball reaches?
m

(c) When the tennis ball is at a height of 2.6 m above the ground, what is its speed?
m/s

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4. 7/10 points  |  Previous Answers WebAssignAlgPhys1 2.8.002. My Notes
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1 2 3 4 5 6 7 8 9 10
1/1 1/1 1/1 1/1 0/1 1/1 1/1 1/1 0/1 0/1
1/50 1/50 1/50 1/50 1/50 1/50 1/50 1/50 1/50 1/50
Total
7/10
 
An athlete is training on a 100 m long linear track. His motion is described by the graph of his position vs. time, below.
WebAssign Plot
(a) For each segment of the graph, find the magnitude 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     Incorrect: Your answer is incorrect. seenKey

1.33
Remember that velocity is the change in position per the change in time. If the graph plots position vs. time, what property of the plot in each segment is related to velocity? Think about the initial and final times and positions for each segment. 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 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     Incorrect: Your answer is incorrect. 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. 1/4 points  |  Previous Answers WebAssignAlgPhys1 4.2.004. My Notes
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1 2 3 4
1/1 0/1 0/1 0/1
1/50 1/50 1/50 1/50
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1/4
 
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?
     Correct: Your answer is correct.

(b) Which object or objects have the largest magnitude of force? (Select all that apply.)
Incorrect: Your answer is incorrect.


Consider the rules for vector addition.
(c) Which object or objects move with constant velocity? (Select all that apply.)
Incorrect: Your answer is incorrect.


What is the basic premise of Newton's First Law?
(d) Which object or objects move with changing speed? (Select all that apply.)
Incorrect: Your answer is incorrect.


What is the basic premise of Newton's First Law?


Solution or Explanation
(a) Diagram (i) is the only case where the two forces are equal in magnitude and opposite in direction.

(b) For Diagram (ii) and Diagram (iii), the resultant force is obtained by considering the net force in the x-direction and the net force in the y-direction. (Assume that the positive x-direction is to the right and the positive y-direction is up.)
For Diagram (ii):
Fx = F cos(45) − F = −0.293F
Fy = −F sin(45) − F = −0.707F
Fnet = 
(−0.293F)2 + (−0.707F)2
 = 0.765F
For Diagram (iii):
Fnet = 
F2 + F2
 = 1.414F
For Diagram (iv):
Fnet = 2F − F = F
Therefore, Diagram (iii) has the largest magnitude of force.

(c) Notice that Diagram (i) is the only one that has a net zero force.

(d) The cases with a net non-zero force will move with changing speed.

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6. 1/3 points  |  Previous Answers WebAssignAlgPhys1 4.4.003. My Notes
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1/3
 
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
Incorrect: Your answer is incorrect.


Consider just the external forces acting on the system of interest to find the reaction forces.
(b) the Moon exerts a force
Correct: Your answer is correct.


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


Consider just the external forces acting on the system of interest to find the reaction forces.




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.
(i) 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.

(ii) Earth exerts a downward gravitational pull on the mower. The reaction force is the mower exerting an upward pull on the Earth.

(iii) 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.
(i) Earth exerts a force to the left on the Moon. The reaction force is the Moon pulling to the right on the Earth.

(ii) 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.
(i) 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.

(ii) 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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–/3
 
A 74-kg man stands on a bathroom scale inside an elevator.
(a) The elevator accelerates upward from rest at a rate of 1.30 m/s2 for 1.50 s. What does the scale read during this 1.50 s interval?
N

(b) The elevator continues upward at constant velocity for 8.50 s. What does the scale read now?
N

(c) While still moving upward, the elevator's speed decreases at a rate of 0.400 m/s2 for 3.00 s. What is the scale reading during this time?
N

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–/4
 
The three diagrams below show a block of mass m being pulled or pushed at constant velocity along a table with a force
P.
 Assume the surfaces to be frictionless.
(a) What is the magnitude of the normal force in each case? Use the following as necessary: g, P, 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. 1/3 points  |  Previous Answers WebAssignAlgPhys1 4.5.023. My Notes
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1/3
 
(a) A block of mass m = 4.30 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?
Correct: Your answer is correct. seenKey

42.1

 N

(b) Two blocks each of mass m = 4.30 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

42.1
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 = 4.30 kg is in equilibrium on an incline plane of angle θ = 31.0° 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?
Incorrect: Your answer is incorrect. seenKey

21.7
Did you draw a free-body diagram and identify all forces acting on the system? Take the x direction to be parallel to the plane and y direction to be perpendicular to the plane. N


Solution or Explanation
(a) The system is in equilibrium. Therefore the net force will equal zero.
WebAssign Plot
T − mg = 0 or T = mg
For a string that has negligible mass and does not stretch, the tension will be the same along every point on the string. The spring scale reads this tension. So the spring scale reading is equal to
mg = (4.30 kg)(9.8 m/s2) = 42.1 N.

(b) The free-body diagram for each mass will be the same as in part (a). Since both strings will have the same tension, the spring scale will read
42.1 N.

(c) The scale will read the tension in the string. Consider the x direction to be parallel to the incline as shown in the free-body diagram.
WebAssign Plot
Fx = 0
mg sin θ − T = 0
T = mg sin θ = (4.30 kg)(9.8 m/s2) sin(31.0°) = 21.7 N

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–/3
 
As shown in the figure below, a box of mass
m = 13.9 kg
 is released from rest (at position A) at the top of a 30.0° frictionless incline. The box slides a distance
d = 4.80 m
 down the incline before it encounters (at position B) a spring and compresses it an amount
xC = 0.200 m
 (to point C) before coming momentarily to rest. Using energy content, determine the following.
(a) speed of the box at position B
vB = m/s


(b) spring constant
k = N/m


(c) the physical quantity that is constant throughout the process
    

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11. –/4 points WebAssignAlgPhys1 7.6.002. My Notes
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–/4
 
A 1.7-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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0/1
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0/1
 
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 = 0.650. Find h such that the block's speed after crossing the rough patch is 3.00 m/s.
Incorrect: Your answer is incorrect. seenKey

0.557
Try dividing the problem into two parts. For the first part, consider the region where only conservative forces do work on the block and for the second part, consider the region where only nonconservative forces do work on the block. Using conservation of energy, can you obtain an expression for the speed of the block at the bottom of the ramp and before it encounters the rough spot? Can you write an expression for the amount of energy lost by the block as it slides across the rough spot? How does the amount of energy lost by the block as it slides across the rough spot compare to its change in kinetic energy? m


Solution or Explanation
For convenience, let's agree to choose the position of the horizontal section of the track as zero gravitational potential energy. As the block slides from its initial position to the point before the rough patch, energy is conserved and we may write:
ΔKE = −ΔPEg or 
mv02
2
 = −(−mgh)
which gives
v02 = 2gh.
 As the block slides across the rough patch, the work done on it by friction is equal to its change in kinetic energy. This allows us to write:
Wf = ΔKE or −μmgd = 
m(vf2 − v02)
2
.
Canceling the mass m and inserting the expression for
v02
 obtain
−μgd = 
(vf2 − 2gh)
2
Solving this for h, obtain:
h = 
vf2
2g
 + μd.
Inserting values, obtain:
h = 
(3.00 m/s)2
2(9.8 m/s2)
 + (0.650)(0.150 m) = 0.557 m

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