This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Al Marzouqi, Khalfan – Homework 5 – Due: Sep 24 2007, 11:00 pm – Inst: Dr C Bradley 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. This is online homework OH05. It is also called HW07(OH05). It is due at 8:00 AM Tuesday, 25 September (UAE time). Since UAE is currently 9 hours ahead of the University of Texas computer, this deadline is also 11:00 PM Monday, 24 September (Texas time). 001 (part 1 of 1) 10 points Twoblocksarearrangedattheendsofamass less string as shown in the figure. The system starts from rest. When the 2 . 69 kg mass has fallen through 0 . 393 m, its downward speed is 1 . 35 m / s. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 69 kg 4 . 19 kg μ a What is the frictional force between the 4 . 19 kg mass and the table? Correct answer: 10 . 4093 N. Explanation: Given : m 1 = 2 . 69 kg , m 2 = 4 . 19 kg , v = 0 m / s , and v = 1 . 35 m / s . Basic Concept: Newton’s Second Law F = M a Solution: The acceleration of m 1 is obtained from the equation v 2 v 2 = 2 a ( s s ) a = v 2 v 2 2 h = (1 . 35 m / s) 2 (0 m / s) 2 2(0 . 393 m) = 2 . 3187 m / s 2 . Consider free body diagrams for the two masses T m 1 g a T N μ N a m 2 g Because m 1 and m 2 are tied together with string, they have same the speed and the same acceleration, so the net force exerted on m 2 is F 2 = m 2 a The net force on m 1 is m 1 a = m 1 g T , so that T = m 1 g m 1 a . Thus F 2 = T f k , f k = T F 2 = m 1 g ( m 1 + m 2 ) a = (2 . 69 kg)(9 . 8 m / s 2 ) (2 . 69 kg + 4 . 19 kg) × (2 . 3187 m / s 2 ) = 10 . 4093 N . keywords: 002 (part 1 of 1) 10 points The magnitude of each force is 221 N, the force on the right is applied at an angle 22 ◦ and the mass of the block is 12 kg. The coefficient of friction is 0 . 184. The acceleration of gravity is 9 . 8 m / s 2 . 12 kg μ = 0 . 184 2 2 1 N 22 ◦ 221 N Al Marzouqi, Khalfan – Homework 5 – Due: Sep 24 2007, 11:00 pm – Inst: Dr C Bradley 2 What is the magnitude of the resulting ac celeration? Correct answer: 34 . 9585 m / s 2 . Explanation: Given : F = 221 N , α = 22 ◦ , μ = 0 . 184 , m = 12 kg , and g = 9 . 8 m / s 2 , m F α N mg F μ N The block is in equilibrium vertically, so X F up = X F down F sin α + N = mg N = mg F sin α The block is accelerating horizontally, so F net = ma = F + F cos α μ N F net = ma = F (1 + cos α + μ sin α ) μmg a = F m ‡ 1 + cos α + μ sin α · μg = 221 N 12 kg × h 1 + cos22 ◦ + (0 . 184)sin22 ◦ i (0 . 184)(9 . 8 m / s 2 ) = 34 . 9585 m / s 2 . keywords: 003 (part 1 of 1) 10 points A 3.90 kg block is pushed along the ceiling with a constant applied force of 81.0 N that acts at an angle of 62.0 ◦ with the horizontal....
View
Full
Document
This note was uploaded on 02/16/2010 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.
 Spring '09
 NAJAFZADEH
 mechanics, Work

Click to edit the document details