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# solution_pdf_courseuserassignment=8640593 -...

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Unformatted text preview: mitchell (jm56986) – HW #4 – montgomery – (24131) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Start a ball rolling down a bowling alley and you’ll find it moves slightly slower with time. Does this violate Newton’s law of inertia? Defend your answer. 1. None of these 2. No; air resistance and friction act upon the ball. correct 3. Yes; no force acts upon it. 4. No; the law of inertia can also be applied to moving objects. 5. Yes; the air resistance cancels the friction and the total force on the ball is zero. Explanation: If there were no force acting on the ball it would continue to move without slowing down. Air resistance along with slight fric- tion with the lane slows the ball down. This doesn’t violate the law of inertia because ex- ternal forces act on the ball. 002 10.0 points What is the acceleration of a 43 kg block of cement when pulled sideways with a net force of 388 N? Correct answer: 9 . 02326 m / s 2 . Explanation: F = ma a = F m = 388 N 43 kg = 9 . 02326 m / s 2 . keywords: 003 10.0 points The Earth exerts an 730 N gravitational force on a man. What force does the man exert on the Earth? Correct answer: 730 N. Explanation: By Newton’s third law, the man exerts an equal but opposite force on the Earth. 004 10.0 points A block is at rest on an inclined plane. 6 8 k g μ s = . 4 3 θ c Find the critical angle, θ c , at which the block just begins to slide. Correct answer: 23 . 2677 ◦ . Explanation: Basic Concepts: Friction: f s ≤ μ s N , f k = μ k N Let : m = 68 kg , μ s = 0 . 43 , θ c = 23 . 2677 ◦ , and v f = final speed . F f N mg 23 . 2677 ◦ Solution: At the critical angle θ c , the mag- nitude of the static frictional force attains its maximum value μ s N . Since the block is at rest, the component of the gravitational force parallel to the incline is equal in magnitude to the frictional force mg sin θ c = f s = μ s N . (1) Also the component of the gravitational force perpendicular to the incline is equal in mag- nitude to the normal force N = mg cos θ c . (2) mitchell (jm56986) – HW #4 – montgomery – (24131) 2 Combining (1) and (2) yields mg sin θ c = μ s mg cos θ c , or tan θ c = μ s θ c = tan − 1 μ s = arctan(0 . 43) = 23 . 2677 ◦ . 005 10.0 points No force is necessary to 1. keep an object moving the way it is al- ready moving. correct 2. cause a change in the motion of an ob- ject. 3. stop an object from moving. 4. start an object moving. Explanation: 006 (part 1 of 4) 10.0 points You have two forces, 92 . 2 N and 69 . 8 N....
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## This note was uploaded on 11/22/2009 for the course PHYS 24131 taught by Professor Montogomery during the Fall '09 term at Northwest MS.

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