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Coletta - Physics Fundamentals 1/e (Homework)

James Finch

Physics - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 24 / 28

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

Last Saved : n/a Saving...  ()

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1 2 3 4 5 6 7 8 9 10 11
1/2 0/1 2/2 7/9 2/2 1/1 3/3 3/3 2/2 1/1 2/2
Total
24/28 (85.7%)

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.

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1. 1/2 points  |  Previous Answers ColFunPhys1 4.P.006. My Notes
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1 2
1/1 0/1
15/50 5/50
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1/2
 
The particle shown below is at rest, where F1 = 6.90 N, and F2 = 10.4 N. Find the magnitude and direction of F.
magnitude Correct: Your answer is correct.
direction ° counterclockwise from the +x-axis

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2. 0/1 points  |  Previous Answers ColFunPhys1 4.P.007. My Notes
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1
0/1
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0/1
 
A boat is pulled at constant velocity by the two forces shown below, where F = 49.0 N, and θ = 11.0°. Find the horizontal force exerted on the boat by the water. The +x direction is to the right.
Incorrect: Your answer is incorrect. in the +x direction.
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3. 2/2 points  |  Previous Answers ColFunPhys1 4.P.015. My Notes
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1 2
1/1 1/1
1/50 1/50
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2/2
 
A 3.00 kg mass is acted upon by four forces in the horizontal (xy) plane, as shown in figure below, where F1 = 36.0 N, F2 = 48.0 N, F3 = 29.0 N, and F4 = 45.0 N. Find the acceleration of the mass.
magnitude Correct: Your answer is correct.
direction Correct: Your answer is correct. ° counterclockwise from the +x-axis

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4. 7/9 points  |  Previous Answers ColFunPhys1 4.WE.004. My Notes
Question Part
Points
Submissions Used
1 2 3 4 5 6 7 8 9
1/1 1/1 1/1 1/1 0/1 –/1 1/1 1/1 1/1
2/50 10/50 4/50 1/50 1/50 0/50 1/50 1/50 1/50
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7/9
 
Example 4: Forces on a Man
Find the forces acting on a standing man whose mass is 104.0 kg.
Part 1 of 5 - Solution
According to Eq. 4-5, the man experiences a force w in the downward Correct: Your answer is correct. seenKey

downward

 direction (the direction of g), and the magnitude of this force is
w = mg = (104.0 kg)(386.4 inches/s^2 Correct: Your answer is correct. seenKey

9.80 m/s^2

 ) = 1.024 kiloN Correct: Your answer is correct. seenKey

1020 N
Part 2 of 5 - Solution
This force can also be expressed in pounds.
w = 1019.2 N 
1.00 lb
4.45 N
 = 229 lb Correct: Your answer is correct. seenKey

229 lb
Part 3 of 5 - Solution
Since the man is standing at rest, his acceleration is zero and so the second law implies that there must be another force to cancel the weight and produce a resultant force equal to zero, as shown in the figure above. This other force is produced by the contact between his feet and the surface on which he is standing. We denote this surface force by S and use the second law to solve for it:
ΣF = ma = 0
S + w = 238.8 Incorrect: Your answer is incorrect. seenKey

0
Thus
S = (No Response) -w
Part 4 of 5 - Solution
The equation above says that the forces are oppositely directed and have equal magnitudes:
S = w = 1020 N Correct: Your answer is correct. seenKey

1020 N
Part 5 of 5 - One Step Further
We could have just as easily solved this problem using Newton's second law in component form. Taking the positive y-axis in the upward direction, we have
ΣFy = may = 0
Thus
$$S−w
Correct: Your answer is correct. S-w = 0
Therefore
S = w = 1020 N Correct: Your answer is correct. seenKey

1020 N
Conclusion
The forces S and w are equal here because the man is stationary. It is possible for him to increase the force S by pushing down on the ground with a force greater than his weight. By Newton's third law, the upward force on his feet will then be greater. There would then be a resultant upward force of magnitude S − w, and the man would accelerate upward. In other words, by pushing on the ground with a force greater than his weight, the man can jump.
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5. 2/2 points  |  Previous Answers ColFunPhys1 4.P.026. My Notes
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1 2
1/1 1/1
1/50 1/50
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2/2
 
The block in the figure below rests on a frictionless surface. Find its instantaneous acceleration when the spring on the left is compressed 4.43 cm while the spring on the right is stretched 10.5 cm. Each spring has a force constant of 1.17 103 N/m.
magnitude Correct: Your answer is correct.
direction Correct: Your answer is correct.

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6. 1/1 points  |  Previous Answers ColFunPhys1 3.A2L.002. My Notes
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1
1/1
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1/1
 
For which cases is the acceleration the same for the motion described in both columns?
Case Column 1 Column 2
(A) A car goes from 0 to 60 mps in 6 s along a straight highway. A car goes from 60 to 0 mph in 6 s along a straight highway.
(B) A race car travels around a circular track at 50 mph. A race car travels around the same circular track at 100 mph.
(C) A ball is thrown straight up. It rises 20 ft. Ignore the effects of the air. A ball is dropped straight down. It falls 20 ft. Ignore the effects of the air.

Correct: Your answer is correct.
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7. 3/3 points  |  Previous Answers ColFunPhys1 4.P.029. My Notes
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1 2 3
1/1 1/1 1/1
2/50 2/50 1/50
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3/3
 
Find the tension in each string in the figure below, where θ1 = 42.5°, θ2 = 20.5°, and w = 28.5 N.
T1 = Correct: Your answer is correct.
T2 = Correct: Your answer is correct.
T3 = Correct: Your answer is correct.

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8. 3/3 points  |  Previous Answers ColFunPhys1 4.P.031. My Notes
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1 2 3
1/1 1/1 1/1
1/50 2/50 2/50
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3/3
 
A crate weighing 5.61 102 N is lifted at a slow, constant speed by ropes attached to the crate at A and B in the figure below, where θ = 38.5°. These two ropes are joined together at point C, and a single vertical rope supports the system.
(a) Find the tension T1 in the vertical rope.
Correct: Your answer is correct.

(b) Find the tensions T2 and T3 in the other ropes.
T2 = Correct: Your answer is correct.
T3 = Correct: Your answer is correct.
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9. 2/2 points  |  Previous Answers ColFunPhys1 4.P.039. My Notes
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1/1 1/1
5/50 1/50
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2/2
 
Two blocks connected by a string are on a horizontal frictionless surface. The blocks are connected to a hanging weight by means of a string that passes over a pulley as shown in the figure below, where m1 = 1.90 kg, m2 = 2.75 kg, and m3 = 5.05 kg.
(a) Find the tension T in the string connecting the two blocks on the horizontal surface.
Correct: Your answer is correct.

(b) How much time is required for the hanging weight to fall 10.0 cm if it starts from rest?
Correct: Your answer is correct.
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10. 1/1 points  |  Previous Answers ColFunPhys1 4.P.044. My Notes
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1
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6/50
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1/1
 
A car is stuck in a mudhole. In order to move the car, the driver attaches one end of a rope to the car and the other end to a tree 11.5 m away, stretching the rope as much as possible as shown in the figure below. The driver then applies a horizontal force of 3.60 102 N perpendicular to the rope at its midpoint. The rope stretches, with its center point moving 51.0 cm to the side as a result of the applied force. The car begins to move slowly. What is the tension in the rope?
Correct: Your answer is correct.
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11. 2/2 points  |  Previous Answers ColFunPhys1 19.CQ.008. My Notes
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1/1 1/1
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2/2
 
A lemon with a copper rod and a steel paperclip stuck in the sides acts as a battery (see figure).
(a) What is the source of the emf?
     Correct: Your answer is correct.

(b) This battery is not at all practical as a source of energy because of its very high internal resistance. What would happen to the voltmeter's reading if the lemon were connected to a 1 Ω resistor to form a complete circuit? (Assume the resistor is connected at the same points as the copper rod and steel paperclip.)
     Correct: Your answer is correct.
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