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# oldhw13 - hinojosa(jlh3938 oldhomework 13 Turner(58185 This...

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hinojosa (jlh3938) – oldhomework 13 – Turner – (58185) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points The suspended 2 . 9 kg mass on the right is moving up, the 2 . 3 kg mass slides down the ramp, and the suspended 7 . 8 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 19 . The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 2 . 3 kg μ = 0 . 19 37 7 . 8 kg 2 . 9 kg What is the acceleration of the three block system? Correct answer: 4 . 47421 m / s 2 . Explanation: Let : m 1 = 2 . 9 kg , m 2 = 2 . 3 kg , m 3 = 7 . 8 kg , and θ = 37 . Basic Concept: F net = m a negationslash = 0 Solution: The acceleration a of each mass is the same, but the tensions in the two strings will be different. Let T 1 be the tension in the right string and T 3 the tension in the left string. Consider the free body diagrams for each mass T 3 m 3 g a T 1 m 1 g a T 3 T 1 N μ N a m 2 g For the mass m 1 , T 1 acts up and the weight m 1 g acts down, with the acceleration a di- rected upward F net 1 = m 1 a = T 1 m 1 g (1) For the mass on the table, the parallel compo- nent of its weight is m g sin θ and the perpen- dicular component of its weight is m g cos θ . ( N = m g cos θ from equilibrium). The accel- eration a is directed down the table, T 3 and the parallel weight component m 2 g sin θ act down the table, and T 1 and the frictional force μ N = μ m 2 g cos θ act up the table F net 2 = m 2 a (2) = T 3 + m 2 g sin θ T 1 μ m 2 g cos θ . For the mass m 3 , T 3 acts up and the weight m 3 g acts down, with the acceleration a di- rected downward F net 3 = m 3 a = m 3 g T 3 . (3) Adding Eqs. (1), (2), & (3) yields ( m 1 + m 2 + m 3 ) a = m 3 g + m 2 g sin θ μ m 2 g cos θ m 1 g . Solving for a , we have a = [ m 2 sin θ μ m 2 cos θ + ( m 3 m 1 )] g m 1 + m 2 + m 3 = (2 . 3 kg) (9 . 8 m / s 2 ) sin 37 2 . 9 kg + 2 . 3 kg + 7 . 8 kg (0 . 19) (2 . 3 kg) (9 . 8 m / s 2 ) cos 37 2 . 9 kg + 2 . 3 kg + 7 . 8 kg + (7 . 8 kg 2 . 9 kg) (9 . 8 m / s 2 ) 2 . 9 kg + 2 . 3 kg + 7 . 8 kg = 4 . 47421 m / s 2 .

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hinojosa (jlh3938) – oldhomework 13 – Turner – (58185) 2 002 (part 2 of 3) 10.0 points What is the tension in the cord connected to the 2 . 9 kg block? Correct answer: 41 . 3952 N. Explanation: Using Eq. 1, we have T 1 = m 1 g + m 1 a = (2 . 9 kg) (9 . 8 m / s 2 + 4 . 47421 m / s 2 ) = 41 . 3952 N . 003 (part 3 of 3) 10.0 points What is the tension in the cord connected to the 7 . 8 kg block? Correct answer: 41 . 5412 N. Explanation: Using Eq. 3, we have T 3 = m 3 g m 3 a = (7 . 8 kg) (9 . 8 m / s 2 4 . 47421 m / s 2 ) = 41 . 5412 N . 004 (part 1 of 2) 10.0 points Two blocks are arranged at the ends of a mass- less cord over a frictionless massless pulley as shown in the figure. Assume the system starts from rest. When the masses have moved a dis- tance of 0 . 437 m, their speed is 1 . 35 m / s.
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