{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

í��ì�¬ì�¤2í��ì��ë£¨ì��

# í��ì�¬ì�¤2í��ì��ë£¨ì�� -...

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
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: min (mm64262) – Homework 2 – lee – (GEDB00864) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A light, inextensible cord passes over a light, frictionless pulley with a radius of 5 . 2 cm. It has a(n) 17 kg mass on the left and a(n) 6 . 8 kg mass on the right, both hanging freely. Initially their center of masses are a vertical distance 2 . 9 m apart. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 9 m 5 . 2 cm ω 17 kg 6 . 8 kg At what rate are the two masses accelerat- ing when they pass each other? Correct answer: 4 . 2 m / s 2 . Explanation: Let : R = 5 . 2 cm , m 1 = 6 . 8 kg , m 2 = 17 kg , h = 2 . 9 m , and v = ω R . Consider the free body diagrams 17 kg 6 . 8 kg T T m 2 g m 1 g a a Since the larger mass will move down and the smaller mass up, we can take motion downward as positive for m 2 and motion up- ward as positive for m 1 . Apply Newton’s second law to m 1 and m 2 respectively and then combine the results: For mass 1: summationdisplay F 1 : T- m 1 g = m 1 a (1) For mass 2: summationdisplay F 2 : m 2 g- T = m 2 a (2) We can add Eqs. (1) and (2) above and obtain: m 2 g- m 1 g = m 1 a + m 2 a a = m 2- m 1 m 1 + m 2 g = 17 kg- 6 . 8 kg 17 kg + 6 . 8 kg (9 . 8 m / s 2 ) = 4 . 2 m / s 2 . T = m 1 ( g + a ) = (6 . 8 kg) (9 . 8 m / s 2 + 4 . 2 m / s 2 ) = 95 . 2 N . 002 (part 1 of 3) 10.0 points A boy throws a ball upward. Compare the magnitudes of the gravitational accelerations at three points along the path of the ball. Point A is before the ball reaches the top. a) Point A on the way up. b) Point B is at the top. c) Point C is after it has passed the top and on the way down. min (mm64262) – Homework 2 – lee – (GEDB00864) 2 B A C The magnitudes of the acceleration are re- lated as 1. a A < a B . 2. a A = g . correct Explanation: The gravitational acceleration near the sur- face of the earth is a constant, so the correct choice is a A = g . 003 (part 2 of 3) 10.0 points The magnitudes of the acceleration are re- lated as 1. a B = 0 . 2. a B > a C . 3. a B = a A . correct Explanation: The gravitational acceleration near the sur- face of the earth is a constant, so the correct choice is a B = a A . 004 (part 3 of 3) 10.0 points The magnitudes of the acceleration are re- lated as 1. a C < a B and a A < a B . 2. a C = a B = a A . correct Explanation: The gravitational acceleration near the sur- face of the earth is a constant, so the correct choice is a C = a B = a A . 005 (part 1 of 2) 10.0 points A block with a mass of 6.3 kg is held in equilibrium on a frictionless incline of 31.0 ◦ by the horizontal force vector F , as shown....
View Full Document

{[ snackBarMessage ]}

### Page1 / 8

í��ì�¬ì�¤2í��ì��ë£¨ì�� -...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online