Quiz2-solutions - Version 043 – Quiz2 – gentle –...

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Unformatted text preview: Version 043 – 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 A block of mass 3 . 45 kg lies on a frictionless horizontal surface. The block is connected by a cord passing over a pulley to another block of mass 3 . 17 kg which hangs in the air, as shown on the following picture. Assume the cord to be light (massless and weightless) and unstretchable and the the pulley to have no friction and no rotational inertia. The acceleration of gravity is 9 . 8 m / s 2 . 3 . 45 kg 3 . 17 kg Calculate the acceleration of the first block. 1. 6.27532 2. 2.76935 3. 4.58163 4. 5.40269 5. 3.1586 6. 6.79583 7. 3.93543 8. 4.69275 9. 4.16628 10. 6.55626 Correct answer: 4 . 69275 m / s 2 . Explanation: Let : m 1 = 3 . 45 kg and m 2 = 3 . 17 kg . m 1 m 2 a T N m 1 g a T m 2 g Since the cord is unstretchable, the first block accelerates to the right at exactly the same rate a as the second (hanging) block ac- celerates downward. Also, the cord’s tension pulls the first block to the right with exactly the same tension T as it pulls the second block upward. The only horizontal force acting on the first block is the cord’s tension T , hence by New- ton’s Second Law m 1 a = F net → 1 = T . The second block feels two vertical forces: The cord’s tension T (upward) and the block’s own weight W 2 = m 2 g (downward). Conse- quently, m 2 a = F net ↓ 2 = m 2 g- T . Adding the two equations together, we arrive at ( m 1 + m 2 ) a = m 2 g , and hence a = m 2 m 1 + m 2 g = 3 . 17 kg 3 . 45 kg + 3 . 17 kg (9 . 8 m / s 2 ) = 4 . 69275 m / s 2 . 002 10.0 points Assume: All surfaces, wheels, and pulley are frictionless. The inextensible cord and pulley are massless. m 1 m 2 F M Given M = 41 . 8 kg, m 1 = 4 . 69 kg, m 2 = 6 . 18 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 Version 043 – Quiz2 – gentle – (56245) 2 blocks to remain stationary relative to the cart? 1. 1099.46 2. 680.151 3. 251.018 4. 113.895 5. 118.649 6. 86.9808 7. 103.421 8. 1190.99 9. 408.39 10. 36.0321 Correct answer: 680 . 151 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 Newton’s 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...
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This note was uploaded on 02/28/2012 for the course PHY 316 taught by Professor Markert during the Fall '10 term at University of Texas at Austin.

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Quiz2-solutions - Version 043 – Quiz2 – gentle –...

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