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Unformatted text preview: mitchell (jm56986) – HW #4 – montgomery – (24131) 1 This printout should have 21 questions. Multiplechoice 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.
 Fall '09
 montogomery
 Physics

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