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DeAntonio-Concepts Of Physics 1/e Sample (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 10 / 43

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

Last Saved : n/a Saving...  ()

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    Questions are available in WebAssign for DeAntonio’s Concepts of Physics, 1st edition, by Michael DeAntonio, published by Great River Learning. These questions come from WebAssign’s exclusive collection of physics questions, authored and peer-reviewed by experienced physics educators.

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1. 0/2 points  |  Previous Answers DeAntCOP1 2.WA.002. My Notes
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1 2
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0/2
 
Find the following 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
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
Incorrect: Your answer is incorrect. seenKey

9
Displacement is the change in position, that is, the difference between the initial and final position. 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 DeAntCOP1 2.WA.022.tut. My Notes
Question Part
Points
Submissions Used
1 2 3
1/1 0/1 –/1
1/5 1/5 0/5
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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 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?
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?
    
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3. –/3 points DeAntCOP1 2.WA.035.tut. My Notes
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1 2 3
–/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 6.0 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 2.2 m above the ground, what is its speed?
m/s
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4. –/10 points DeAntCOP1 2.WA.049. My Notes
Question Part
Points
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1 2 3 4 5 6 7 8 9 10
–/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1 –/1
0/5 0/5 0/5 0/5 0/5 0/5 0/5 0/5 0/5 0/5
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–/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 and direction of the athlete's velocity.
magnitude vA     m/s
direction vA    
magnitude vB     m/s
direction vB    
magnitude vC     m/s
direction vC    
magnitude vD     m/s
direction vD    

(b) What are the magnitude and direction of the athlete's average velocity over the entire 60 s interval?
magnitude     m/s
direction    
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5. 4/4 points  |  Previous Answers DeAntCOP1 3.WA.004. My Notes
Question Part
Points
Submissions Used
1 2 3 4
1/1 1/1 1/1 1/1
1/5 1/5 1/5 1/5
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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.
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(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.)
Correct: Your answer is correct.


(c) Which object or objects move with constant velocity? (Select all that apply.)
Correct: Your answer is correct.


(d) Which object or objects move with changing speed? (Select all that apply.)
Correct: Your answer is correct.



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 DeAntCOP1 3.WA.008. My Notes
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1 2 3
0/1 1/1 0/1
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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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7. –/3 points DeAntCOP1 3.WA.037.tut. My Notes
Question Part
Points
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1 2 3
–/1 –/1 –/1
0/5 0/5 0/5
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–/3
 
A 66-kg man stands on a bathroom scale inside an elevator.
(a) The elevator accelerates upward from rest at a rate of 1.20 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 0.450 m/s2 for 3.00 s. What is the scale reading during this time?
N
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8. 4/4 points  |  Previous Answers DeAntCOP1 3.WA.042. My Notes
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1 2 3 4
1/1 1/1 1/1 1/1
1/5 1/5 1/5 1/5
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4/4
 
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 = 
mg
Correct: Your answer is correct. m g
case (ii)    N = 
mg+Fsin(θ)
Correct: Your answer is correct. m g + F sin(theta)
case (iii)    N = 
mg−Fsin(θ)
Correct: Your answer is correct. m g - F sin(theta)

(b) How would your answer to part (a) change if, all else being the same, the object moved with constant acceleration?
     Correct: Your answer is correct.


Solution or Explanation
(a) The free body diagram for each situation is shown below.
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To find the normal force, we consider the net force in the vertical direction. In each case Fy = 0.
case (i)    N − mg = 0
N = mg
case (ii)    N − mg − F sin θ = 0
N = mg + F sin θ
case (iii)    N − mg + F sin θ = 0
N = mg − F sin θ

(b) The forces in the vertical direction will not change and therefore the normal force in each case will be the same as in part (a).
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9. –/3 points DeAntCOP1 3.WA.059. My Notes
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–/3
 
(a) A block of mass m = 5.10 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?
N

(b) Two blocks each of mass m = 5.10 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?
N

(c) A block of mass m = 5.10 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?
N
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10. –/3 points DeAntCOP1 4.WA.044.tut. My Notes
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1 2 3
–/1 –/1 –/1
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–/3
 
As shown in the figure below, a box of mass
m = 13.2 kg
 is released from rest (at position A) at the top of a 30.0° frictionless incline. The box slides a distance
d = 3.40 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 DeAntCOP1 4.WA.057. My Notes
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–/4
 
A 1.9-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 DeAntCOP1 4.WA.070. My Notes
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1
–/1
0/5
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–/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 = 0.510. Find h such that the block's speed after crossing the rough patch is 3.70 m/s.
m
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