This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: quiz 01 BAUTISTA, ALDO Due: Feb 15 2006, 10:00 pm 1 Version number encoded for clicker entry: V1:1, V2:2, V3:1, V4:1, V5:1. Question 1 Part 1 of 1. 10 points. A ball on the end of a string is whirled around in a horizontal circle of radius 0 . 345 m. The plane of the circle is 1 . 82 m above the ground. The string breaks and the ball lands 2 . 26 m away from the point on the ground directly beneath the balls location when the string breaks. The acceleration of gravity is 9 . 8 m / s 2 . Find the centripetal acceleration of the ball during its circular motion. 1. 33 . 9499 m / s 2 2. 35 . 1241 m / s 2 3. 36 . 224 m / s 2 4. 37 . 4209 m / s 2 5. 38 . 6627 m / s 2 6. 39 . 8586 m / s 2 correct Explanation: In order to find the centripetal acceleration of the ball, we need to find the initial velocity of the ball. Let y be the distance above the ground. After the string breaks, the ball has no initial velocity in the vertical direction, so the time spent in the air may be deduced from the kinematic equation, y = 1 2 g t 2 . Solving for t , t = r 2 y g . Let d be the distance traveled by the ball. Then v x = d t = d r 2 y g . Hence, the centripetal acceleration of the ball during its circular motion is a c = v 2 x r = d 2 g 2 y r = 39 . 8586 m / s 2 . Question 2 Part 1 of 2. 10 points. A book is at rest on an incline as shown above. A hand, in contact with the top of the book, produces a constant force F hand vertically downward. F hand B o o k The following figures show several attempts at drawing freebody diagrams for the book. Which figure has the correct directions for each force? Note: The magnitude of the forces are not necessarily drawn to scale. 1. weight force normal friction 2. weight friction force normal 3. weight friction normal force quiz 01 BAUTISTA, ALDO Due: Feb 15 2006, 10:00 pm 2 4. weight force normal friction 5. weight normal friction force 6. weight force friction normal correct Explanation: The normal force points perpendicular to the surface of the inclined plane. The weight force points down. The F hand also points down. The friction force keeps the book from sliding and consequently points up the incline. Question 3 Part 2 of 2. 10 points. For the normal force exerted on the book by the wedge in the diagram, which force(s) complete(s) the force pair for Newtons third law (actionreaction)? 1. the pull of the earth on the book 2. the pull of the book on the earth 3. the sum of the component of gravity per pendicular to the surface of the incline and the component of F hand perpendicular to the surface of the incline 4. the normal force exerted on the wedge by the book correct 5. the component of F hand pointing perpen dicular to the surface of the incline 6. the component of gravity pointing parallel to the surface of the incline Explanation: The force that completes the third law pair with the normal force of the wedge on the book is the normal force of the book on the wedge....
View
Full
Document
 Spring '08
 Turner

Click to edit the document details