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: 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
 Spring '09
 NAJAFZADEH
 mechanics, Work

Click to edit the document details