CP Physics Name___Answer Key______Review - MP 2 Quarterly Date ______
Quarterly: Tuesday, January 26 ISN Pages 37-66
Know the following terms:
Acceleration due to Gravity (g) Centripetal Acceleration
Centripetal Force Cosine
Force Frequency
Friction g-Force
Inertia Magnitude
Mass Newton’s First Law
Newton’s Second Law Newton’s Third Law
Newton’s Law of Universal Gravitation Period
Resultant Sine
Tangent Vector
Weight
Know what each of the following equations is used for and what the units for each term is.
W = mg �!"" � − ����� = �
�!"# = �� �! = � �! 2 � � � = � = 2��� � 1 1 � = � = � � �! � = �! = � �! ! � ��!�! �� � = � = �! �! �� � = � 1
1. In the absence of a net force, a moving object will A. slow down and eventually stop B. stop immediately C. turn right D. move with constant velocity E. turn left
2. In order for a rocket ship in deep space, far from any other objects, to move in a straight line with constant speed it must exert a net force that is A. proportional to its mass B. proportional to its weight C. proportional to its velocity D. zero E. proportional to its displacement
3. A boy rides a bicycle at a constant velocity. Which of the following about the net force is true? A. There is a net force acting in the velocity direction B. There is a net force acting opposite to the velocity direction C. The net force is zero D. There is a net force acting perpendicularly to the velocity direction E. None from the above
4. The acceleration of an object is proportional to A. the net force acting on it B. its position C. its velocity D. its mass E. its displacement
5. The acceleration of an object is inversely proportional to A. the net force acting on it B. its position C. its velocity D. its mass E. its displacement
6. Mass and weight A. Both have the same measuring units B. Both have different measuring units C. Both represent force of gravity D. Both represent measure of inertia E. None from the above
7. Which of the following is an example of a force which acts at a distance (without contact)? A. Tension B. Gravity C. Static friction D. Kinetic friction E. Normal force
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8. An elevator of mass M is pulled upwards by a cable; the elevator is moving up but accelerating in the opposite directon. What is the tension in the cable (neglecting the mass of the cable)? A. less than zero B. between zero and Mg (although it is moving up it is accelerating DOWN) C. equal to Mg D. greater than Mg E. zero
9. A box is being pushed by a constant force along a horizontal surface. If the object’s velocity is constant, we can infer that there is ______acting on the box A. a frictional force (must be the same magnitude of the applied force so net force = 0) B. a net downward force C. no frictional force D. a net force upward E. a net force in the acceleration direction
10. A car moves around a circular path of a constant radius at a constant speed. Which of the following statements is true? A. The car’s velocity is constant B. The car’s acceleration is constant C. The car’s acceleration is zero D. The car’s velocity is directed toward the center E. The car’s acceleration is directed toward the center
11. An object moves around a circular path at a constant speed and makes five complete revolutions in 20 seconds. What is the period of rotation? A. 5 s B. 10 s C. 4 s D. 20 s E. 15 s
12. An object moves around a circular path at a constant speed and makes ten complete revolutions in 5 seconds. What is the frequency of rotation? A. 2 Hz B. 4 Hz C. 6 Hz D. 10 Hz E. 20 Hz
13. An object rotates with a period of 10 s. How many revolutions will it make in 25 s? A. 10 B. 15 C. 5 D. 2.5 E. 2
14. An object rotates with a frequency of 300Hz. How many revolutions will it make in 15 s? A. 1000 B. 1500 C. 2000 D. 3500 E. 4500
15. An object rotates with a period of 0.5 s. What is the frequency of rotations? A. 1.0 Hz B. 1.5 Hz C. 2.0 Hz D. 2.5 Hz E. 3.0 Hz
16. An object rotates with a frequency of 50 Hz. What is the period of rotations A. 0.02 s B. 0.15 s C. 0.25 s D. 0.05 s E. 0.03 s
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17. An apple hangs on a tree on the Earth. Which of the following statements is correct? A. The force on the apple is greater than the force on the Earth because the Earth is more massive. B. The force on the Earth is greater than the force on the apple because the Earth is more massive. C. The force on the apple is less than the force on the Earth because the tree is supporting the apple. D. The forces on the apple and Earth are equal. (Force of Gravity and remember Newton’s Third Law?) E. The forces on the apple and the Earth are zero because they are not touching each other.
Problems – Complete on a separate sheet of paper.
18. If a net horizontal force of 175 N is applied to a bike whose mass is 43 kg what acceleration is produced?
�!"# = �� � 175 � � � = !"# = = �. � � 43 �� ��
19. The magnitude of a boy’s weight is 270 N. What is his mass?
� = �� � −270 � � = = � = ��. � �� � −9.8 �!
20. Find the weight of a 60 kg table.
� � = �� = 60 �� −9.8 = −��� � �!
21. A train with a mass of 25000 kg increases its speed from 10 m/s to 25 m/s in 20 seconds. Assume that the acceleration is constant and that you can neglect friction. a. Find the acceleration of the train
� � ∆� 25 − 10 � � = = � � = �. �� ∆� 20 � ��
b. Find the distance traveled during this 20 s?
! ! �! = �! + 2�∆� � � ! ! (25 )! − (10 )! �! − �! � � ∆� = = � = ��� � 2� 2 (0.75 ) �!
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c. Draw a free- body diagram for the train; Assume that the applied force is to the right and the train is moving to the right. Therefore, friction would be to the left (opposing the motion).
Ff Fapp
d. Find the average net force supplied by the locomotive.
� � = �� = 25,000 �� 0.75 = ��, ��� � !"# �!
22. A 1500 kg elevator moves up and down on a cable. Calculate the tension in the cable for the following cases: ! � = �� = 1,500 �� −9.8 = −��, ��� � (������ �� ��� ��������) Fg !!
a. The elevator moves at constant speed upward Free-body diagram is to the right. The forces are equal because it is moving at constant velocity. FT
FT = -Fg = -(-14,700N) = 14,700 N Fg
b. The elevator moves at constant speed downward
Free-body diagram is to the right. The forces are equal because it is moving at constant velocity. FT
FT = -Fg = -(-14,700N) = 14,700 N Fg
c. The elevator accelerates upward at a constant rate of 1.2 m/s2 Net force is pointed upward because accelerating upward. Thus, the force of tension Is greater than the weight of the elevator.
� F � = �� = 1,500 �� 1.2 = 1,800 � T !"# �! �!"# = �! + �!
�! = �!"# − �! = 1,800 � − −14,700 � = ��, ��� � Fg
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d. The elevator accelerates downward at a constant rate of 1.2 m/s2
Net force is pointed downward because accelerating downward (so it is a negative number). Thus, the force of tension is less than the weight of the elevator. FT � � = �� = 1,500 �� −1.2 = −1,800 � !"# �! �!"# = �! + �! Fg �! = �!"# − �! = −1,800 � − −14,700 � = ��, ��� �
23. An object is experiencing an acceleration of 12 m/s2 while traveling in a circle at a velocity of 3.1 m/s. What is the radius of its motion? �! � = ! �
� ! (3.1 )! � � � = = � = �. � � �! 12 �!
24. A 61 kg object is experiencing a net force of 25 N while traveling in a circle of radius 35 m. What is its velocity?
�! = ��! � 25 � � � = ! = = 0.409 ! � 61 �� �!
�! � = ! � � � � = (� )(�) = 0.409 (35�) = �. � ! �! �
25. Compute g for the surface of the sun. Its radius is 7.0 x108 m and its mass is 2.0 x 1030 kg.
��! 6.67 � 10!!! (2.0 � 10!"��) �� ��! � � = = = ��� �! (7.0 � 10! �)! ��
26. Compute the velocity of an object orbiting at a distance of 8.5 x 109m from the center of a spherical object whose mass is 5.0 x 1028 kg.
��! 6.67 � 10!!! (5.0 � 10!"��) �� ��! � � = = = ��, ��� � (8.5 � 10! �) �
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A. Compute the orbital period of that object.
2�� � = � 2�� 2� (8.5 � 10!�) � = = � = �, ���, ��� � = ��� ����� = �� ���� � 19,808 �
27. As shown in the diagram below, a 5.0 kg space rock is located 2.5x107 m from the center of the earth. The mass of the earth is 6.0x1024 kg.
a
v
Earth Space Rock
a. Determine the force of gravity acting on the space rock, due to the earth. Calculate the magnitude and state the direction.
��! 6.67 � 10!!! (6.0 � 10!"��)(5.0 ��) �� � ��! � = ! ! = = �. � � �! (2.5 � 10! �)!
b. Compare your answer in a) to the force of gravity acting on the earth, due to the space rock. Indicate that force on the diagram above.
The space rock is pulling the Earth to right with a force of 3.2 N. The forces are the same because of Newton’s Third Law.
c. On the diagram above, indicate the direction the space rock would accelerate if released. Label that vector “a”.
d. Calculate the acceleration the rock would experience.
�!"# = �� � 3.2 � � � = !"# = = �. �� � 5 �� ��
e. **If instead of falling, the object were in a stable orbit, indicate on the diagram above a possible direction of its velocity. Label that vector “v”.
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f. **Calculate the velocity the rock needs to be in a stable orbit.
To be in a stable orbit the centripetal acceleration must be equal to the acceleration due to gravity at that point.
��! 6.67 � 10!!! (6.0 � 10!!��) �� ��! � � = = = �, ��� � (2.5 � 10! �) �
g. **Calculate the period of the rock orbiting the earth.
2�� � = � 2�� 2� (2.5 � 10!�) � = = � = ��, ��� � = �� ����� � 4,001 �
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