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**ConcepTest Clicker Questions**

Chapter 6 Physics, 4th Edition James S. Walker Copyright © 2010 Pearson Education, Inc.

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Question 6.1a Tension I You tie a rope to a tree and you pull on the rope with a force of 100 N. What is the tension in the rope? a) 0 N b) 50 N c) N d) N e) N Answer: c

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Question 6.1a Tension I You tie a rope to a tree and you pull on the rope with a force of 100 N. What is the tension in the rope? a) 0 N b) 50 N c) N d) N e) N The tension in the rope is the force that the rope “feels” across any section of it (or that you would feel if you replaced a piece of the rope). Because you are pulling with a force of 100 N, that is the tension in the rope.

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Question 6.1b Tension II Two tug-of-war opponents each pull with a force of 100 N on opposite ends of a rope. What is the tension in the rope? a) 0 N b) 50 N c) N d) N e) N Answer: c

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Question 6.1b Tension II Two tug-of-war opponents each pull with a force of 100 N on opposite ends of a rope. What is the tension in the rope? a) 0 N b) 50 N c) N d) N e) N This is literally the identical situation to the previous question. The tension is not 200 N !! Whether the other end of the rope is pulled by a person, or pulled by a tree, the tension in the rope is still 100 N !!

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Question 6.1c Tension III You and a friend can each pull with a force of 20 N. If you want to rip a rope in half, what is the best way? a) you and your friend each pull on opposite ends of the rope b) tie the rope to a tree, and you both pull from the same end c) it doesn’t matter—both of the above are equivalent d) get a large dog to bite the rope Answer: b

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Question 6.1c Tension III You and a friend can each pull with a force of 20 N. If you want to rip a rope in half, what is the best way? a) you and your friend each pull on opposite ends of the rope b) tie the rope to a tree, and you both pull from the same end c) it doesn’t matter—both of the above are equivalent d) get a large dog to bite the rope Take advantage of the fact that the tree can pull with almost any force (until it falls down, that is!). You and your friend should team up on one end, and let the tree make the effort on the other end.

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Question Three Blocks Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tension in each of the strings? a) T1 > T2 > T3 b) T1 < T2 < T3 c) T1 = T2 = T3 d) all tensions are zero e) tensions are random T3 T2 T1 3m 2m m a Answer: a

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Question Three Blocks Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. What is the relationship between the tension in each of the strings? a) T1 > T2 > T3 b) T1 < T2 < T3 c) T1 = T2 = T3 d) all tensions are zero e) tensions are random T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass. a T3 3m T2 T1 2m m Follow-up: What is T1 in terms of m and a?

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**Question 6.3 Over the Edge m m a a F = 98 N Case (1) Case (2)**

a) case (1) b) acceleration is zero c) both cases are the same d) depends on value of m e) case (2) In which case does block m experience a larger acceleration? In case (1) there is a 10 kg mass hanging from a rope and falling. In case (2) a hand is providing a constant downward force of 98 N. Assume massless ropes. m a F = 98 N m Answer: e a 10 kg Case (1) Case (2)

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**Question 6.3 Over the Edge m m a a F = 98 N Case (1) Case (2)**

a) case (1) b) acceleration is zero c) both cases are the same d) depends on value of m e) case (2) In which case does block m experience a larger acceleration? In case (1) there is a 10 kg mass hanging from a rope and falling. In case (2) a hand is providing a constant downward force of 98 N. Assume massless ropes. In case (2) the tension is 98 N due to the hand. In case (1) the tension is less than 98 N because the block is accelerating down. Only if the block were at rest would the tension be equal to 98 N. m a F = 98 N m a 10 kg Case (1) Case (2)

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**Question 6.4 Friction a) the force from the rushing air pushed it off**

A box sits in a pickup truck on a frictionless truck bed. When the truck accelerates forward, the box slides off the back of the truck because: a) the force from the rushing air pushed it off b) the force of friction pushed it off c) no net force acted on the box d) truck went into reverse by accident e) none of the above Answer: c

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**Question 6.4 Friction a) the force from the rushing air pushed it off**

A box sits in a pickup truck on a frictionless truck bed. When the truck accelerates forward, the box slides off the back of the truck because: a) the force from the rushing air pushed it off b) the force of friction pushed it off c) no net force acted on the box d) truck went into reverse by accident e) none of the above Generally, the reason that the box in the truck bed would move with the truck is due to friction between the box and the bed. If there is no friction, there is no force to push the box along, and it remains at rest. The truck accelerated away, essentially leaving the box behind!!

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**Question 6.5 Antilock Brakes**

Antilock brakes keep the car wheels from locking and skidding during a sudden stop. Why does this help slow the car down? a) mk > ms so sliding friction is better b) mk > ms so static friction is better c) ms > mk so sliding friction is better d) ms > mk so static friction is better e) none of the above Answer: d

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**Question 6.5 Antilock Brakes**

Antilock brakes keep the car wheels from locking and skidding during a sudden stop. Why does this help slow the car down? a) mk > ms so sliding friction is better b) mk > ms so static friction is better c) ms > mk so sliding friction is better d) ms > mk so static friction is better e) 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.

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**Question 6.6 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? a) pushing her from behind b) pulling her from the front c) both are equivalent d) it is impossible to move the sled e) tell her to get out and walk 1 2 Answer: b

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**Question 6.6 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? a) pushing her from behind b) pulling her from the front c) both are equivalent d) it is impossible to move the sled e) 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

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**Question 6.7 Will It Budge? T m a) moves to the left**

b) moves to the right c) moves up d) moves down e) 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? Answer: e T m Static friction (ms = 0.4 )

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**Question 6.7 Will It Budge? T m a) moves to the left**

b) moves to the right c) moves up d) moves down e) 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?

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**Question 6.8a 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? a) component of the gravity force parallel to the plane increased b) coefficient of static friction decreased c) normal force exerted by the board decreased d) both #1 and #3 e) all of #1, #2, and #3 Answer: d Net Force Normal Weight

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**Question 6.8a Sliding Down I**

a) component of the gravity force parallel to the plane increased b) coefficient of static friction decreased c) normal force exerted by the board decreased d) both #1 and #3 e) all of #1, #2, and #3 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? 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). Because 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

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**Question 6.8b Sliding Down II**

A mass m is placed on an inclined plane (m > 0) and slides down the plane with constant speed. If a similar block (same m) of mass 2m were placed on the same incline, it would: a) not move at all b) slide a bit, slow down, then stop c) accelerate down the incline d) slide down at constant speed e) slide up at constant speed m Answer: d

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**Question 6.8b Sliding Down II**

A mass m is placed on an inclined plane (m > 0) and slides down the plane with constant speed. If a similar block (same m) of mass 2m were placed on the same incline, it would: a) not move at all b) slide a bit, slow down, then stop c) accelerate down the incline d) slide down at constant speed e) slide up at constant speed q W N f Wx Wy The component of gravity acting down the plane is double for 2m. However, the normal force (and hence the friction force) is also double (the same factor!). This means the two forces still cancel to give a net force of zero.

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Question Tetherball In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point? a) toward the top of the pole b) toward the ground c) along the horizontal component of the tension force d) along the vertical component of the tension force e) tangential to the circle W T Answer: c

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Question Tetherball In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point? a) toward the top of the pole b) toward the ground c) along the horizontal component of the tension force d) along the vertical component of the tension force e) 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

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**Question 6.10a Around the Curve I**

You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn. From your perspective in the car, what do you feel is happening to you? a) you are thrown to the right b) you feel no particular change c) you are thrown to the left d) you are thrown to the ceiling e) you are thrown to the floor Answer: a

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**Question 6.10a Around the Curve I**

You are a passenger in a car, not wearing a seat belt. The car makes a sharp left turn. From your perspective in the car, what do you feel is happening to you? a) you are thrown to the right b) you feel no particular change c) you are thrown to the left d) you are thrown to the ceiling e) you are thrown to the floor The passenger has the tendency to continue moving in a straight line. From your perspective in the car, it feels like you are being thrown to the right, hitting the passenger door.

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**Question 6.10b Around the Curve II**

a) centrifugal force is pushing you into the door b) the door is exerting a leftward force on you c) both of the above d) neither of the above During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening? Answer: b

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**Question 6.10b Around the Curve II**

a) centrifugal force is pushing you into the door b) the door is exerting a leftward force on you c) both of the above d) neither of the above During that sharp left turn, you found yourself hitting the passenger door. What is the correct description of what is actually happening? The passenger has the tendency to continue moving in a straight line. There is a centripetal force, provided by the door, that forces the passenger into a circular path.

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**Question 6.10c Around the Curve III**

a) car’s engine is not strong enough to keep the car from being pushed out b) friction between tires and road is not strong enough to keep car in a circle c) car is too heavy to make the turn d) a deer caused you to skid e) none of the above You drive your dad’s car too fast around a curve and the car starts to skid. What is the correct description of this situation? Answer: b

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**Question 6.10c Around the Curve III**

a) car’s engine is not strong enough to keep the car from being pushed out b) friction between tires and road is not strong enough to keep car in a circle c) car is too heavy to make the turn d) a deer caused you to skid e) none of the above You drive your dad’s car too fast around a curve and the car starts to skid. What is the correct description of this situation? The friction force between tires and road provides the centripetal force that keeps the car moving in a circle. If this force is too small, the car continues in a straight line! Follow-up: What could be done to the road or car to prevent skidding?

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**d e c a b Question 6.11 Missing Link**

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? Answer: b

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**d e a c b Question 6.11 Missing Link**

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? e 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?

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**Question 6.12 Ball and String**

a) T2 = ¼T1 b) T2 = ½T1 c) T2 = T1 d) T2 = 2T1 e) T2 = 4T1 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 Answer: d 2

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**Question 6.12 Ball and String**

a) T2 = ¼T1 b) T2 = ½T1 c) T2 = T1 d) T2 = 2T1 e) T2 = 4T1 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

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**Question 6.13 Barrel of Fun a b c d e**

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? a b c d e Answer: a

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**Question 6.13 Barrel of Fun a b c d e**

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? a b c d e 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?

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**Question 6.14a Going in Circles I**

You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by your seat is equal to your weight mg. How does N change at the top of the Ferris wheel when you are in motion? a) N remains equal to mg b) N is smaller than mg c) N is larger than mg d) none of the above Answer: b

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**Question 6.14a Going in Circles I**

You’re on a Ferris wheel moving in a vertical circle. When the Ferris wheel is at rest, the normal force N exerted by your seat is equal to your weight mg. How does N change at the top of the Ferris wheel when you are in motion? a) N remains equal to mg b) N is smaller than mg c) N is larger than mg d) none of the above You are in circular motion, so there has to be a centripetal force pointing inward. At the top, the only two forces are mg (down) and N (up), so N must be smaller than mg. Follow-up: Where is N larger than mg?

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**Question 6.14b Going in Circles II**

a) Fc = N + mg b) Fc = mg – N c) Fc = T + N – mg d) Fc = N e) Fc = mg A skier goes over a small round hill with radius R. Because 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 Answer: b

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**Question 6.14b Going in Circles II**

a) Fc = N + mg b) Fc = mg – N c) Fc = T + N – mg d) Fc = N e) Fc = mg A skier goes over a small round hill with radius R. Because 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?

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**Question 6.14c Going in Circles III**

You swing a ball at the end of string in a vertical circle. Because 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? a) Fc = T – mg b) Fc = T + N – mg c) Fc = T + mg d) Fc = T e) Fc = mg R v top Answer: c

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**Question 6.14c Going in Circles III**

You swing a ball at the end of string in a vertical circle. Because 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? a) Fc = T – mg b) Fc = T + N – mg c) Fc = T + mg d) Fc = T e) 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?

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