1. ConcepTest Clicker Questions Chapter 5
Physics: for Scientists & Engineers with Modern Physics, 4th edition
Giancoli
© 2008 Pearson Prentice Hall
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2. ConcepTest 5.2 Antilock Brakes
Antilock brakes keep the car wheels from locking and skidding during a sudden stop. Why does this help slow the car down?
1) mk > ms so sliding friction is better
2) mk > ms so static friction is better
3) ms > mk so sliding friction is better
4) ms > mk so static friction is better
5) none of the above

3. ConcepTest 5.2 Antilock Brakes
Antilock brakes keep the car wheels from locking and skidding during a sudden stop. Why does this help slow the car down?
1) mk > ms so sliding friction is better
2) mk > ms so static friction is better
3) ms > mk so sliding friction is better
4) ms > mk so static friction is better
5) none of the above
Static friction is greater than sliding friction, so by keeping the wheels from skidding, the static friction force will help slow the car down more efficiently than the sliding friction that occurs during a skid.

4. ConcepTest 5.3 Going Sledding
Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this?
1) pushing her from behind
2) pulling her from the front
3) both are equivalent
4) it is impossible to move the sled
5) tell her to get out and walk 1 2

5. ConcepTest 5.3 Going Sledding
Your little sister wants you to give her a ride on her sled. On level ground, what is the easiest way to accomplish this?
1) pushing her from behind
2) pulling her from the front
3) both are equivalent
4) it is impossible to move the sled
5) tell her to get out and walk
In Case 1, the force F is pushing down (in addition to mg), so the normal force is larger. In Case 2, the force F is pulling up, against gravity, so the normal force is lessened. Recall that the frictional force is proportional to the normal force. 1 2

6. ConcepTest 5.4 Will it Budge?
1) moves to the left
2) moves to the right
3) moves up
4) moves down
5) the box does not move
A box of weight 100 N is at rest on a floor where ms = A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move? T m
Static friction (ms = 0.4 )

7. ConcepTest 5.4 Will it Budge?
1) moves to the left
2) moves to the right
3) moves up
4) moves down
5) the box does not move
A box of weight 100 N is at rest on a floor where ms = A rope is attached to the box and pulled horizontally with tension T = 30 N. Which way does the box move?
The static friction force has a maximum of msN = 40 N. The tension in the rope is only 30 N. So the pulling force is not big enough to overcome friction. T m
Static friction (ms = 0.4 )
Follow-up: What happens if the tension is 35 N? What about 45 N?

8. ConcepTest 5.5a Sliding Down I
A box sits on a flat board. You lift one end of the board, making an angle with the floor. As you increase the angle, the box will eventually begin to slide down. Why?
1) component of the gravity force parallel to the plane increased
2) coeff. of static friction decreased
3) normal force exerted by the board decreased
4) both #1 and #3
5) all of #1, #2 and #3
Net Force
Normal
Weight

9. ConcepTest 5.5a Sliding Down I
A box sits on a flat board. You lift one end of the board, making an angle with the floor. As you increase the angle, the box will eventually begin to slide down. Why?
1) component of the gravity force parallel to the plane increased
2) coeff. of static friction decreased
3) normal force exerted by the board decreased
4) both #1 and #3
5) all of #1, #2 and #3
As the angle increases, the component of weight parallel to the plane increases and the component perpendicular to the plane decreases (and so does the Normal force). Since friction depends on Normal force, we see that the friction force gets smaller and the force pulling the box down the plane gets bigger.
Net Force
Normal
Weight

10. ConcepTest 5.6 Tetherball
In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point?
1) toward the top of the pole
2) toward the ground
3) along the horizontal component of the tension force
4) along the vertical component of the tension force
5) tangential to the circle W T

11. ConcepTest 5.6 Tetherball
In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point?
1) toward the top of the pole
2) toward the ground
3) along the horizontal component of the tension force
4) along the vertical component of the tension force
5) tangential to the circle W T
The vertical component of the tension balances the weight. The horizontal component of tension provides the centripetal force that points toward the center of the circle. W T

12. ConcepTest 5.8 Missing Link1 2 3 5 4
A ping pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the ping pong ball leaves the track, which path will it follow?

13. ConcepTest 5.8 Missing Link1 2 3 5 4
A ping pong ball is shot into a circular tube that is lying flat (horizontal) on a tabletop. When the ping pong ball leaves the track, which path will it follow?
Once the ball leaves the tube, there is no longer a force to keep it going in a circle. Therefore, it simply continues in a straight line, as Newton’s First Law requires!
Follow-up: What physical force provides the centripetal force?

14. ConcepTest 5.9 Ball and String
Two equal-mass rocks tied to strings are whirled in horizontal circles. The radius of circle 2 is twice that of circle 1. If the period of motion is the same for both rocks, what is the tension in cord 2 compared to cord 1?
1) T2 = 1/4 T1
2) T2 = 1/2 T1
3) T2 = T1
4) T2 = 2 T1
5) T2 = 4 T1 1 2

15. ConcepTest 5.9 Ball and String
1) T2 = 1/4 T1
2) T2 = 1/2 T1
3) T2 = T1
4) T2 = 2 T1
5) T2 = 4 T1
Two equal-mass rocks tied to strings are whirled in horizontal circles. The radius of circle 2 is twice that of circle 1. If the period of motion is the same for both rocks, what is the tension in cord 2 compared to cord 1? 1
The centripetal force in this case is given by the tension, so T = mv2/r. For the same period, we find that v2 = 2v1 (and this term is squared). However, for the denominator, we see that r2 = 2r1 which gives us the relation T2 = 2T1. 2

16. ConcepTest 5.10 Barrel of Fun
A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? 1 2 3 4 5

17. ConcepTest 5.10 Barrel of Fun
A rider in a “barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her? 1 2 3 4 5
The normal force of the wall on the rider provides the centripetal force needed to keep her going around in a circle. The downward force of gravity is balanced by the upward frictional force on her, so she does not slip vertically.
Follow-up: What happens if the rotation of the ride slows down?

18. ConcepTest 5.11b Going in Circles II
1) Fc = N + mg
2) Fc = mg – N
3) Fc = T + N – mg
4) Fc = N
5) Fc = mg
A skier goes over a small round hill with radius R. Since she is in circular motion, there has to be a centripetal force. At the top of the hill, what is Fc of the skier equal to? R v

19. ConcepTest 5.11b Going in Circles II
1) Fc = N + mg
2) Fc = mg – N
3) Fc = T + N – mg
4) Fc = N
5) Fc = mg
A skier goes over a small round hill with radius R. Since she is in circular motion, there has to be a centripetal force. At the top of the hill, what is Fc of the skier equal to? v
Fc points toward the center of the circle, i.e., downward in this case. The weight vector points down and the normal force (exerted by the hill) points up. The magnitude of the net force, therefore, is: Fc = mg – N mg N R
Follow-up: What happens when the skier goes into a small dip?

20. ConcepTest 5.11c Going in Circles III
You swing a ball at the end of string in a vertical circle. Since the ball is in circular motion there has to be a centripetal force. At the top of the ball’s path, what is Fc equal to?
1) Fc = T – mg
2) Fc = T + N – mg
3) Fc = T + mg
4) Fc = T
5) Fc = mg R v
top

21. ConcepTest 5.11c Going in Circles III
You swing a ball at the end of string in a vertical circle. Since the ball is in circular motion there has to be a centripetal force. At the top of the ball’s path, what is Fc equal to?
1) Fc = T – mg
2) Fc = T + N – mg
3) Fc = T + mg
4) Fc = T
5) Fc = mg
Fc points toward the center of the circle, i.e. downward in this case. The weight vector points down and the tension (exerted by the string) also points down. The magnitude of the net force, therefore, is: Fc = T + mg v T mg R
Follow-up: What is Fc at the bottom of the ball’s path?