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# solution_pdf - Version 058 FIRST EXAM Kleinman(58835 This...

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Version 058 – FIRST EXAM – Kleinman – (58835) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The horizontal surface on which the objects slide is frictionless. The acceleration of gravity is 9 . 8 m / s 2 . 1 kg 8 kg 6 kg F F r If F = 15 N and F r = 7 N, what is the magnitude of the force exerted on the block with mass 8 kg by the block with mass 6 kg? 1. 6.94118 2. 7.8 3. 6.35294 4. 10.2 5. 7.33333 6. 13.0 7. 9.33333 8. 11.6471 9. 10.0 10. 11.4 Correct answer: 10 . 2 N. Explanation: m 1 m 2 m 3 F F r Given : vector F = +15 N ˆ ı , vector F r = 7 N ˆ ı , m 1 = 1 kg , m 2 = 8 kg , m 3 = 6 kg , and g = 9 . 8 m / s 2 . Note: F is acting on the combined mass of the three blocks, resulting in a common acceleration after accounting for friction. m 1 F 21 F m 2 F 32 F 12 m 3 F r F 23 Let F , F r , F 32 represent the force exerted on the system from the right, from the left, and the force exerted on m 2 by m 3 , respec- tively. Note: vector F 21 = vector F 12 and vector F 32 = vector F 23 , and bardbl vector F 21 bardbl = bardbl vector F 12 bardbl and bardbl vector F 32 bardbl = bardbl vector F 23 bardbl , where bardbl vector F bardbl ≡ F is the magnitude of vector F . The sum of the forces acting on each block separately are m 1 a = + F F 21 = + F F 12 (1) m 2 a = + F 12 F 32 = + F 12 F 23 (2) m 3 a = + F 23 F r (3) To find the acceleration we can treat the three masses as a single object or add the forces acting on each component of the sys- tem, Eqs. 1, 2, and 3. F F r = ( m 1 + m 2 + m 3 ) a Solving for a, we have a = F F r m 1 + m 2 + m 3 (4) = 15 N 7 N 1 kg + 8 kg + 6 kg = 0 . 533333 m / s 2 . We can solve for F 23 using Eq. 3 and sub- stituting a from Eq. 4. The result is F 23 = m 3 a + F r (3) = m 3 ( F F r ) m 1 + m 2 + m 3 + F r ( m 1 + m 2 + m 3 ) m 1 + m 2 + m 3 = m 3 F ( m 1 + m 2 ) F r m 1 + m 2 + m 3 (5) = (6 kg) (15 N) 1 kg + 8 kg + 6 kg + (1 kg + 8 kg) (7 N) 1 kg + 8 kg + 6 kg = 10 . 2 N .

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Version 058 – FIRST EXAM – Kleinman – (58835) 2 Alternative Solution: Using Eq. 3 and the numerical result of Eq. 4, we have F 23 = m 3 a + F r (3) = (6 kg) (0 . 533333 m / s 2 ) + 7 N = 10 . 2 N .
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