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solution9_pdf

# solution9_pdf - arellano(aa39398 Practice Homework Chapter...

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arellano (aa39398) – Practice Homework Chapter 5 Solutions – Weathers – (17101) 1 This print-out should have 65 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 8) 10.0 points Assume you are on a planet similar to Earth where the acceleration of gravity is approxi- mately 10 m / s 2 and the positive directions for displacement, velocity, and acceleration are upward. At time t = 0 s, an elevator is at a displace- ment of x = 0 m with a velocity of v = 0 m / s. A student whose normal weight is 400 N stands on a scale in an elevator and records the scale reading as a function of time. The data are shown in the graph. 0 5 10 15 20 0 200 400 600 800 time (s) Scale Reading (N) In which direction does the force due to the scale act on the student? 1. backward 2. right 3. up correct 4. The net force on the student is zero, so the force the scale exerts on the student has no direction. 5. forward 6. down 7. left Explanation: The scale exerts a normal force (Scale Read- ing F s ), opposite to the force of gravity m g ; i.e. , upward. 002 (part 2 of 8) 10.0 points What is the mass of the student? 1. m 70 kg 2. m 30 kg 3. m 10 kg 4. m 60 kg 5. m 40 kg correct 6. m 80 kg 7. m 50 kg 8. m 90 kg 9. m 20 kg Explanation: Since W = m g , we have m = W g = 400 N 10 m / s 2 = 40 kg . 003 (part 3 of 8) 10.0 points Calculate the acceleration of the elevator dur- ing the fourth 5 s interval (from 15 s to 20 s). Correct answer: 5 m / s 2 . Explanation: The net force F net on the student is F net = F s 400 N = m a , so a = F s 400 N m . The accelerations during the 5-second in- tervals are a 0 5 = 400 N 400 N 40 kg = 0 m / s 2

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arellano (aa39398) – Practice Homework Chapter 5 Solutions – Weathers – (17101) 2 a 5 10 = 500 N 400 N 40 kg = 2 . 5 m / s 2 a 10 15 = 400 N 400 N 40 kg = 0 m / s 2 a 15 20 = 600 N 400 N 40 kg = 5 m / s 2 . 004 (part 4 of 8) 10.0 points Choose the correct plot for acceleration vs time. 1. 0 5 10 15 20 10 5 0 5 10 time (s) Acceleration (m/s 2 ) 2. 0 5 10 15 20 10 5 0 5 10 time (s) Acceleration (m/s 2 ) correct 3. 0 5 10 15 20 10 5 0 5 10 time (s) Acceleration (m/s 2 ) 4. 0 5 10 15 20 10 5 0 5 10 time (s) Acceleration (m/s 2 ) 5. 0 5 10 15 20 10 5 0 5 10 time (s) Acceleration (m/s 2 ) 6. 0 5 10 15 20 10 5 0 5 10 time (s) Acceleration (m/s 2 ) 7. 0 5 10 15 20 10 5 0 5 10 time (s) Acceleration (m/s 2 ) 8. 0 5 10 15 20 10 5 0 5 10 time (s) Acceleration (m/s 2 ) Explanation: Use the results calculated in Part 2 to con- struct the acceleration graph: 0 5 10 15 20 10 5 0 5 10 time (s) Acceleration (m/s 2 )
arellano (aa39398) – Practice Homework Chapter 5 Solutions – Weathers – (17101) 3 005 (part 5 of 8) 10.0 points What is the velocity of the elevator at the end of the fourth 5 s interval (at 20 s)? Correct answer: 37 . 5 m / s. Explanation: v = v 0 + a t At the end of five second intervals, we have v 1 = v 0 + a 1 t = 0 m / s + (0 m / s 2 ) (5 s) = 0 m / s v 2 = v 1 + a 2 t = 0 m / s + (2 . 5 m / s 2 ) (5 s) = 12 . 5 m / s v 3 = v 2 + a 3 t = 12 . 5 m / s + (0 m / s 2 ) (5 s) = 12 . 5 m / s v 4 = v 3 + a 3 t = 12 . 5 m / s + (5 m / s 2 ) (5 s) = 37 . 5 m / s .

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