physics HW4 - rose (kmr2427) HW #4 Antoniewicz (57380) 1...

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Unformatted text preview: rose (kmr2427) HW #4 Antoniewicz (57380) 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 The horizontal surface on which the objects slide is frictionless. The acceleration of gravity is 9 . 8 m / s 2 . 3 kg 8 kg 5 kg F F r If F = 23 N and F r = 3 N, what is the magnitude of the force exerted on the block with mass 8 kg by the block with mass 5 kg? Correct answer: 9 . 25 N. Explanation: m 1 m 2 m 3 F F r Given : vector F = +23 N , vector F r = 3 N , m 1 = 3 kg , m 2 = 8 kg , m 3 = 5 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) = 23 N 3 N 3 kg + 8 kg + 5 kg = 1 . 25 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) = (5 kg) (23 N) 3 kg + 8 kg + 5 kg + (3 kg + 8 kg) (3 N) 3 kg + 8 kg + 5 kg = 9 . 25 N . Alternative Solution: Using Eq. 3 and the numerical result of Eq. 4, we have F 23 = m 3 a + F r (3) = (5 kg) (1 . 25 m / s 2 ) + 3 N = 9 . 25 N . 002 (part 1 of 3) 4.0 points rose (kmr2427) HW #4 Antoniewicz (57380) 2 Two forces, 350 N at 12 and 292 N at 32 are applied to a car in an effort to accelerate it. 3311 kg 3 5 N 12 2 9 2 N 32 What is the magnitude of the resultant of these two forces? Correct answer: 595 . 648 N. Explanation: Given : m = 3311 kg , F 1 = 350 N , 1 = 12 , F 2 = 292 N , and 2 = 32 . Consider the side forces: F 1 ,y = F 1 sin 1 F 2 ,y = F 2 sin 2 F y,net = (350 N) sin12 + (292 N) sin( 32 ) = 81 . 9673 N Consider the forward forces: F 1 ,x = F 1 cos 1 F 2 ,x = F 2 cos 2 F x,net = (350 N) cos12 + (292 N) cos( 32 ) = 589 . 982 N Thus the net force is F net = radicalBig (589 . 982 N) 2 + ( 81 . 9673 N) 2 = 595 . 648 N...
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physics HW4 - rose (kmr2427) HW #4 Antoniewicz (57380) 1...

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