HW 12B Dynamics Newton\'s 2nd Law-problems PDF

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sanchez fernande (es43796) – HW 12B Dynamics: Newton’s 2nd Law – wieman – (053305072122) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A 30 kg chair initially at rest on a horizontal floor requires a 164 N horizontal force to set it in motion. Once the chair is in motion, a 129 N horizontal force keeps it moving at a constant velocity. The acceleration of gravity is 9.81 m/s2 . a) What is the coefficient of static friction between the chair and the floor? 002 (part 2 of 2) 10.0 points b) What is the coefficient of kinetic friction between the chair and the floor? 003 (part 1 of 3) 10.0 points Consider a force pulling 3 blocks along a rough horizontal surface, where the masses are a multiple of a given mass m, as shown in the figure below. The coefficient of the kinetic friction is µ. The blocks are pulled by a force of 45 µ m g . 3m

T1

T2

5m

7m

F

µ Determine the acceleration. 1. a = 2 µ g 2. a = 9 µ g 3. a = 6 µ g 4. a = 8 µ g 5. a = 5 µ g 6. a = 7 µ g 7. a = 4 µ g 8. a = µ g 9. a = 3 µ g 004 (part 2 of 3) 10.0 points The equation of motion for the central mass 5 m is given by 1. T2 − T1 − 5 µ m g − 3 µ m g = 5 m a 2. T2 − T1 − 5 µ m g = 5 m a

3. T2 + 5 µ m g = 5 m a 4. T2 − T1 + 5 µ m g = 5 m a 5. T2 + T1 + 5 µ m g − 3 µ m g = 5 m a 6. T2 + T1 − 5 µ m g = 5 m a 7. T2 − 5 µ m g = 5 m a 8. T2 + T1 − 5 µ m g − 3 µ m g = 5 m a 9. T2 + T1 + 5 µ m g = 5 m a 005 (part 3 of 3) 10.0 points Find the tension T1 in the rope between the masses 3 m and 5 m in terms of F . F 1. T1 = 3 2. T1 = F F 3. T1 = 7 F 4. T1 = 2 F 5. T1 = 6 F 6. T1 = 8 F 7. T1 = 4 F 8. T1 = 9 F 9. T1 = 5 006 (part 1 of 3) 10.0 points A hockey puck is hit on a frozen lake and starts moving with a speed of 11.1 m/s. Five seconds later, its speed is 8 m/s. What is its average acceleration? The acceleration of gravity is 9.8 m/s2 . Answer in units of m/s2 . 007 (part 2 of 3) 10.0 points What is the average value of the coefficient of kinetic friction between puck and ice? 008 (part 3 of 3) 10.0 points How far does the puck travel during this 5 s interval? Answer in units of m. 009 10.0 points Block A (66 N) and block B (20 N) are connected by a massless cord and are at rest. Given µs = 0.8.

sanchez fernande (es43796) – HW 12B Dynamics: Newton’s 2nd Law – wieman – (053305072122) 2 The friction on A is : A µs

B Answer in units of N. 010 (part 1 of 3) 10.0 points The blocks are moving. Given: g = 9.8 m/s2 , and two masses 5 kg , and 18 kg . The coefficients of kinetic friction are µ1 = 0.19, µ2 = 0.21, and the force F = 137 N.

The coefficient of static friction between the tires of a car and a horizontal road is 0.25. If the net force on the car is the force of static friction exerted by the road, what is the maximum acceleration of the car when it is braked? The acceleration due to gravity is 9.81 m/s2 . Answer in units of m/s2 . 014 (part 2 of 2) 10.0 points What is the shortest distance in which the car can stop if it is initially traveling at 45 m/s? Answer in units of m. 015 10.0 points Denote the force exerted on block 2 by the block 1 to be F21 . F

137 N

5 kg µ1 = 0.19

18 kg µ2 = 0.21

Find f1, the magnitude of the friction force on block m1 . Answer in units of N. 011 (part 2 of 3) 10.0 points Find a, the magnitude of acceleration of these two blocks. Answer in units of m/s2 . 012 (part 3 of 3) 10.0 points Find the magnitude of the force F21 that the 5 kg mass exerts on the 18 kg mass. 1. F21 = F − µ2 (18 kg) g 2. F21 = F − (5 kg) a − µ2 (18 kg) g 3. F21 = F − (5 kg) a 4. F21 = µ1 (5 kg) g 5. F21 = F − (18 kg) a 6. F21 = (5 kg) a 7. F21 = F − µ1 (5 kg) g − (5 kg) a 8. F21 = F + (5 kg) a − µ1 (5 kg) g 9. F21 = F 10. F21 = F − µ1 (5 kg) g 013 (part 1 of 2) 10.0 points

m1 µ1

m2 µ2

If the acceleration is a, the equation of motion for block m2 is given by 1. F21 − µ2 m2 g = m2 a 2. F − µ1 m1 g − µ2 m2 g = m2 a 3. F − F21 − µ2 m2 g = m2 a 4. F + F21 − µ2 m2 g = m2 a 5. F21 − µ1 m1 g − µ2 m2 g = m2 a...