hw10 - alexander (jra2623) homework 10 Turner (92510) 1...

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alexander (jra2623) – homework 10 – Turner – (92510) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points A snowboarder oF mass m is at rest on the top oF a sand dune. The coe±cients oF static and kinetic Friction are μ s and μ k , respectively. The acceleration oF gravity is 9 . 8 m / s 2 . m μ s k θ What is the angle θ the incline must exceed so that the snowboarder starts sliding? 1. θ = 1 arctan μ k 2. θ = arctan μ k 3. θ = arctan μ s correct 4. θ = arcsin μ s 5. θ = arccos μ s 6. θ = 1 arctan μ s 7. θ = arcsin μ k 8. θ = arccos μ k Explanation: Consider the Free body diagram For the block m g sin θ N = cos m g We know that the largest possible value the static Friction Force can have is F f,max = μ s N , where the normal Force is N = m g cos θ . Thus, since F f = m g sin θ m g sin θ = μ s m g cos θ , and then tan θ = μ s θ = arctan μ s . 002 (part 2 oF 2) 10.0 points What is the acceleration down the incline iF θ exceeds the value Found in the previous question? 1. a = g μ s sin θ 2. a = g ( μ k sin θ cos θ ) 3. a = g ( μ k sin θ μ s cosθ ) 4. a = g μ k sin θ 5. a = g (sin θ μ k cos θ ) correct 6. a = g μ s cos θ 7. a = g ( μ s sin θ cos θ ) 8. a = g ( μ s sin θ μ k cosθ ) 9. a = g μ k cos θ 10. a = g (sin θ μ s cos θ ) Explanation: When θ exceeds the value Found in Part 2, the snowboarder starts moving and the Fric- tion Force is the kinetic Friction F k = μ k N = μ k m g cos θ . Newton’s equation For the snow- boarder reads then m a = m g sin θ F f = m g sin θ μ k m g cos θ , and then a = g (sin θ μ k cos θ ) . 003 (part 1 oF 2) 10.0 points
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alexander (jra2623) – homework 10 – Turner – (92510) 2 A fast-moving VW Beetle traveling at 54 mph hit a mosquito hovering at rest above the road. Which bug experienced the largest force? 1.
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hw10 - alexander (jra2623) homework 10 Turner (92510) 1...

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