HW10_Solutions - hiarker(srh959 homework 10 Turner(58220...

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hiarker (srh959) – homework 10 – Turner – (58220) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points In a certain system of units, 38 . 5 thumbs = 1 meter, 1 pause = 43 . 4 s, and 1 glob = 17 . 1 kg. The unit of force in this system is a perk (p). How many kiloperks (kp) are in 42 . 6 N? Correct answer: 180 . 657 kp. Explanation: The unit of measure for force in the new system is 1 perk = 1 glob · thumb pause 2 . f = parenleftbigg 42 . 6 kg · m s 2 parenrightbigg · 1 glob 17 . 1 kg · 38 . 5 thumbs 1 m · parenleftbigg 43 . 4 s 1 pause parenrightbigg 2 · 1 kp 1000 p = 180 . 657 kp . 002 10.0 points A person weighing 0 . 5 kN rides in an elevator that has a downward acceleration of 1 . 7 m / s 2 . The acceleration of gravity is 9 . 8 m / s 2 . What is the magnitude of the force of the elevator floor on the person? Correct answer: 0 . 413265 kN. Explanation: Basic Concepts: summationdisplay vector F = mvectora vector W = mvectorg Solution: Since W = m g , F net = ma = W − f f = W − ma = W parenleftbigg 1 a g parenrightbigg = (0 . 5 kN) parenleftbigg 1 1 . 7 m / s 2 9 . 8 m / s 2 parenrightbigg = 0 . 413265 kN . 003 (part 1 of 2) 10.0 points A pulley is massless and frictionless. 1 kg, 1 kg, and 11 kg masses are suspended as in the figure. 3 . 8 m 21 . 7 cm ω 1 kg 1 kg 11 kg T 2 T 1 T 3 What is the tension T 1 in the string be- tween the two blocks on the left-hand side of the pulley? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 16 . 5846 N. Explanation: Let : R = 21 . 7 cm , m 1 = 1 kg , m 2 = 1 kg , m 3 = 11 kg , and h = 3 . 8 m . Consider the free body diagrams 1 kg 1 kg 11 kg T 1 T 2 T 3 m 1 g T 1 m 2 g m 3 g a a For each mass in the system vector F net = mvectora .
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hiarker (srh959) – homework 10 – Turner – (58220) 2 Since the string changes direction around the pulley, the forces due to the tensions T 2 and T 3 are in the same direction (up). The acceleration of the system will be down to the right ( m 3 > m 1 + m 2 ), and each mass in the system accelerates at the same rate (the string does not stretch). Let this acceleration rate be a and the tension over the pulley be T T 2 = T 3 . For the lower left-hand mass m 1 the accel- eration is up and T 1 m 1 g = m 1 a . (1) For the upper left-hand mass m 2 the acceler- ation is up and T T 1 m 2 g = m 2 a . (2) For the right-hand mass m 3 the acceleration is down and T + m 3 g = m 3 a . (3) Adding Eqs. (1), (2), and (3), we have ( m 3 m 1 m 2 ) g = ( m 1 + m 2 + m 3 ) a (4) a = m 3 m 1 m 2 m 1 + m 2 + m 3 g (5) = 11 kg 1 kg
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