Quiz2-solutions - Version 044 Quiz2 gentle (56245) 1 This...

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Unformatted text preview: Version 044 Quiz2 gentle (56245) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points Given: 1 hp = 746 W and 1 m / s = 2 . 237 mi / h. When an automobile moves with constant velocity, the power developed in a certain en- gine is 79 hp. What total frictional force acts on the car at v = 60 . 1 mi / h? 1. 1398.52 2. 2194.01 3. 3561.39 4. 3077.27 5. 2894.12 6. 3899.23 7. 2764.79 8. 3467.24 9. 1559.57 10. 4235.59 Correct answer: 2194 . 01 N. Explanation: P = W t = Fd t = Fv Thus F = P v = 79 hp 60 . 1 mi / h = 2194 . 01 N . 002 10.0 points A box is dropped onto a conveyor belt moving at 1 . 7 m / s. The acceleration of gravity is 9 . 8 m / s 2 . If the coefficient of friction between the box and the belt is 0 . 7, what distance does the box move before it moves without slipping? 1. 0.337372 2. 0.920918 3. 0.306973 4. 2.46939 5. 0.291545 6. 0.354308 7. 1.34949 8. 1.23469 9. 0.210641 10. 0.206633 Correct answer: 0 . 210641 m. Explanation: The friction force exerted on the box is f = mg , so the acceleration of the box is a = f m = g . The time that the box needs to reach the speed of conveyor is t = v a = v g . Finally, the distance is d = 1 2 at 2 = 1 2 (6 . 86 m / s 2 ) (0 . 247813 s) 2 = 0 . 210641 m . Alternative Solution: Using v 2 = v 2 + 2 ad, we have d = v 2 2 a = v 2 2 g = (1 . 7 m / s) 2 2 (0 . 7) (9 . 8 m / s 2 ) = 0 . 210641 m . 003 10.0 points Assume: All surfaces, wheels, and pulley are frictionless. The inextensible cord and pulley are massless. Version 044 Quiz2 gentle (56245) 2 m 1 m 2 F M Given M = 27 . 7 kg, m 1 = 2 . 05 kg, m 2 = 5 . 55 kg, and g = 9 . 8 m / s 2 . What horizontal force must be applied to the cart shown in the figure in order for the blocks to remain stationary relative to the cart? 1. 251.018 2. 39.6606 3. 1613.1 4. 936.569 5. 1307.39 6. 658.832 7. 86.9808 8. 1176.63 9. 773.585 10. 986.567 Correct answer: 936 . 569 N. Explanation: Note: The blocks m 1 and m 2 being sta- tionary relative to the cart M means that they have the same non-zero horizontal accel- eration a relative to the ground. Consider the free-body diagrams. N 1 m 1 g T m 2 g T N 2 Applying Newtons second law to the hori- zontal motion of the m 1 block yields m 1 : summationdisplay F x = T = m 1 a, (1) while for the vertical motion of the m 2 block we have m 2 : summationdisplay F y = T- m 2 g = 0 (2) because it accelerates horizontally but not vertically. Combining the above two Eqs. (1) & (2), and solving for the a yields a = m 2 m 1 g . (3) On the other hand, a is related to the ex- ternal force F via the equation of horizontal motion for the whole cart-plus-two-blocks sys- tem: N total ( M + m 1 + m 2 ) g F In light of the above whole-system diagram, the horizontal equation is simply summationdisplay F x = F = ( M + m 1 + m 2 ) a (4) and hence F = [ M + m 1 + m 2 ] bracketleftbigg m 2 m 1 bracketrightbigg...
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Quiz2-solutions - Version 044 Quiz2 gentle (56245) 1 This...

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