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OpenStax - College Physics 2/e (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 15 / 28

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

Last Saved : n/a Saving...  ()

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1/1 2/3 1/2 2/2 3/3 1/3 –/6 –/1 0/2 5/5
Total
15/28 (53.6%)

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1. 1/1 points  |  Previous Answers OSColPhys2 7.5.P.024. My Notes
Question Part
Points
Submissions Used
1
1/1
3/100
Total
1/1
 
A 55.0 kg skier with an initial speed of 11.0 m/s coasts up a 2.50 m high rise as shown in the following figure.
Find her final speed at the top (in m/s), given that the coefficient of friction between her skis and the snow is 0.0800. (Hint: Find the distance traveled up the incline assuming a straight-line path as shown in the figure.)
Correct: Your answer is correct. seenKey

8.15

 m/s
†

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2. 2/3 points  |  Previous Answers OSColPhys2 5.1.P.013. My Notes
Question Part
Points
Submissions Used
1 2 3
1/1 1/1 0/1
5/100 2/100 5/100
Total
2/3
 
Calculate the maximum deceleration (in m/s2) of a car that is heading down a 8° slope (one that makes an angle of 8° with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the static coefficient of friction is involved—that is, the tires are not allowed to slip during the deceleration.
(a)
on dry concrete
Correct: Your answer is correct. seenKey

8.34

 m/s2
(b)
on wet concrete
Correct: Your answer is correct. seenKey

5.43

 m/s2
(c)
on ice, assuming that µs = 0.100, the same as for shoes on ice
Incorrect: Your answer is incorrect. seenKey

-0.393

 m/s2
†

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3. 1/2 points  |  Previous Answers OSColPhys2 5.1.P.019. My Notes
Question Part
Points
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1 2
0/1 1/1
3/100 2/100
Total
1/2
 
A contestant in a winter games event pulls a 30.0 kg block of ice across a frozen lake with a rope over his shoulder as shown in the figure.
The coefficient of static friction is 0.1 and the coefficient of kinetic friction is 0.03.
(a)
Calculate the minimum force F (in N) he must exert to get the block moving.
Incorrect: Your answer is incorrect. seenKey

31

 N
(b)
What is its acceleration (in m/s2) once it starts to move, if that force is maintained?
Correct: Your answer is correct. seenKey

0.655

 m/s2
†

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4. 2/2 points  |  Previous Answers OSColPhys2 18.4.P.028. My Notes
Question Part
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1 2
1/1 1/1
1/100 1/100
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2/2
 
What is the magnitude (in N) and direction of the force exerted on a 3.80 µC charge by a 230 N/C electric field that points due east?
magnitude Correct: Your answer is correct. seenKey

0.000874

 N direction Correct: Your answer is correct. seenKey

due east
†

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5. 3/3 points  |  Previous Answers OSColPhys2 5.1.WA.012.Tutorial. My Notes
Question Part
Points
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1 2 3
1/1 1/1 1/1
2/100 2/100 2/100
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3/3
 
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 = 8.5 N. Here A has a mass mA = 10.2 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 0.02. (Assume
FP
 acts in the +x direction.)
(a)
What is the magnitude (in m/s2) of the acceleration of the two boxes?
Correct: Your answer is correct. seenKey

0.298

 m/s2
(b)
What is the force exerted on mB by mA? In other words what is the magnitude (in N) of the contact force
FC?
Correct: Your answer is correct. seenKey

3.46

 N
(c)
If Alex were to push from the other side on the 7.0-kg box, what would the new magnitude (in N) of
FC
 be?
Correct: Your answer is correct. seenKey

5.04

 N


Solution or Explanation
Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values.
The free-body diagram for each system is shown below.
N
 is the normal force exerted by the surface on the system,
FC
 is the contact force,
f
 is the friction force and
W
 refers to the weight.
WebAssign Plot
WebAssign Plot
WebAssign Plot
Note that for the system containing both boxes, the contact force
FC
 is not external to the system. And when you consider just box B, the force
FP
 does not act on it.
(a)
To find the acceleration of the two boxes, consider both boxes as the system. Apply Newton's Second Law
Fnet = ma
 to this system for motion in the horizontal and vertical directions.
vertical direction:
Fnet, y = may
N − (mA + mB)g = 0
Recall that f = μN and therefore
fAB = μ(mA + mB)g
 = (0.02)(10.2 kg + 7.0 kg)(9.8 m/s2)
 = 3.37 N.
horizontal direction:
Fnet, x = max
FP − fAB = (mA + mB)a
a = 
8.5 N − 3.37 N
(10.2 kg + 7.0 kg)
 = 0.298 m/s2
(b)
To find the contact force FC exerted on box B by box A, consider box B as the system. Now the friction force depends on the normal force exerted by the ground on box B alone.
fB = μmBg = (0.02)(7.0 kg)(9.8 m/s2) = 1.37 N
Apply Newton's Second Law to box B.
FC − fB = mBa
FC = 1.37 N + (7.0 kg)(0.298 m/s2)
 = 3.46 N
(c)
If Alex now pushed on box B rather than on box A, you would have to switch the two masses in the solutions above. Therefore the contact force would now be larger since mass of A is larger. We can check this out by performing the calculations.
fA = μmAg = (0.02)(10.2 kg)(9.8 m/s2) = 2.00 N
FC = 2.00 N + (10.2 kg)(0.298 m/s2) = 5.04 N

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6. 1/3 points  |  Previous Answers OSColPhys2 7.5.WA.041. My Notes
Question Part
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1 2 3
0/1 1/1 –/1
1/100 1/100 0/100
Total
1/3
 
In the figure below, the two boxes are initially moving with a speed v.
(a)
Ignore friction and determine an expression for the distance d the boxes travel before coming to rest. Choose the floor as the position for zero gravitational potential energy. (Use the following as necessary: m1, m2, v, and g for the acceleration due to gravity.)
d =
v22g​
Incorrect: Your answer is incorrect. \frac{(m_1 + m_2)v^2}{2 m_2 g}
Why is energy conserved for the system (both boxes) when there is a nonconservsative force (tension) acting on each box? Can you write a statement of conservation of energy for the system? Can you obtain an expression for the change in kinetic energy and the change in potential energy of the system?
(b)
Is the work done on box 2 by the rope positive, negative, or zero?
     Correct: Your answer is correct.
(c)
Determine an expression for the work done on box 1 by the rope. (Use the following as necessary: m1, m2, v, and g for the acceleration due to gravity.)
W =

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7. –/6 points OSColPhys2 9.4.WA.027. My Notes
Question Part
Points
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1 2 3 4 5 6
–/1 –/1 –/1 –/1 –/1 –/1
0/100 0/100 0/100 0/100 0/100 0/100
Total
–/6
 
A uniform ladder stands on a rough floor and rests against a frictionless wall as shown in the figure.
On the xy coordinate plane there is a ladder leaning against a wall with a man standing near the top of it. The ladder reaches a distance d up on the wall. The man is a distance labeled b to the right of the base of the ladder. The base of the ladder has two arrows extending from it, an arrow pointing up labeled N_1 and an arrow pointing right labeled f_1. The man has an arrow pointing down from his feet labeled with the expression m g. The top of the ladder has a horizontal arrow pointing to the left out of the wall labeled N_2.
Since the floor is rough, it exerts both a normal force
N1
 and a frictional force
f1
 on the ladder. However, since the wall is frictionless, it exerts only a normal force
N2
 on the ladder. The ladder has a length of
L = 4.9 m,
 a weight of
WL = 61.5 N,
 and rests against the wall a distance
d = 3.75 m
 above the floor. If a person with a mass of
m = 90 kg
 is standing on the ladder, determine the following.
(a) the forces exerted on the ladder when the person is halfway up the ladder (Enter the magnitude only.)
N1 = N
N2 = N
f1 = N

(b) the forces exerted on the ladder when the person is three-fourths of the way up the ladder (Enter the magnitude only.)
N1 = N
N2 = N
f1 = N

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8. –/1 points OSColPhys2 11.6.WA.024.Tutorial. My Notes
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–/1
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–/1
 
A rectangular parallelepiped with a square base
d = 0.250 m
 on a side and a height
h = 0.120 m
 has a mass
m = 7.10 kg.
 While this object is floating in water, oil with a mass density
ρo = 710 kg/m3
 is carefully poured on top of the water until the situation looks like that shown in the figure. Determine the height of the parallelepiped in the water.
m
A clear cup is partially filled with water, and has oil floating on top of the water, almost filling the vessel. Completely submerged in the fluids is a block. The block has width and depth labeled d, and a larger height labeled h. The block floats so that part of it is in the water and part of it is in the oil. The portion of the height of the block in the water is labeled h_w, and the portion of the height of the block in the oil is labeled h_o.

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9. 0/2 points  |  Previous Answers OSColPhys2 18.5.WA.056.Tutorial. My Notes
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0/2
 
Newer automobiles have filters that remove fine particles from exhaust gases. This is done by charging the particles and separating them with a strong electric field. Consider a positively charged particle +5.3 µC that enters an electric field with strength
4 ✕ 106 N/C.
 The particle is traveling at 21 m/s and has a mass of
10−9 g.
What is the acceleration of the particle? (Enter the magnitude only.)
m/s2

What is the direction of the acceleration of the particle relative to the electric field?
     Incorrect: Your answer is incorrect.

How is the direction of the electric field related to the direction of the force when the charge is positive?

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10. 5/5 points  |  Previous Answers OSColPhys2 10.IVV.001. My Notes
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1 2 3 4 5
1/1 1/1 1/1 1/1 1/1
1/100 1/100 1/100 1/100 1/100
Total
5/5
 
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?
     Correct: Your answer is correct.
(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?
     Correct: Your answer is correct.
(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.


Solution or Explanation
The two bullets have the same initial momentum, directed upward. They have the same mass, and because they are fired from identically prepared cartridges from the same rifle, they have the same velocity before entering the block.
The two blocks both have the same mass and are both initially at rest before the collision.
Upon impact, conservation of momentum states the momentum of the bullet before impact will be equal to the momentum of the bullet–block system after impact. So both systems have the same upward momentum after impact. And because they have the same mass, they therefore have the same upward velocity after impact.
Because they have the same upward (center-of-mass) velocity after the collision, the two systems then travel to the same maximum height (by constant acceleration kinematics).
The two bullets have the same kinetic energy before the collision, because they have the same mass and speed. Conservation of energy states that both bullet–block systems have the same total energy after the collision.
However, the energy takes different forms after the collision in the two cases. They both do have the same translational kinetic energy after the collision (because they have the same momentum, mass, and center-of-mass speed). But in the first case, more energy has gone into deforming the block, and the bullet is embedded deeper. In the second case, less energy goes into deforming the block—the edge of the block "moves with" the bullet and the bullet does not get embedded as much. The remaining energy instead goes into the rotational kinetic energy of the system.

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