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Unformatted text preview: brown (twb493) – midterm 02 practice – turner – (57340) 1 This printout should have 22 questions. Multiplechoice 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 . 7 kg mass on the right is moving up, the 1 . 9 kg mass slides down the ramp, and the suspended 8 . 3 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 14 . The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 1 . 9 k g μ = . 1 4 32 ◦ 8 . 3 kg 2 . 7 kg What is the acceleration of the three block system? Correct answer: 4 . 84778 m / s 2 . Explanation: Let : m 1 = 2 . 7 kg , m 2 = 1 . 9 kg , m 3 = 8 . 3 kg , and θ = 32 ◦ . Basic Concept: F net = ma 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 mg sin θ and the perpen dicular component of its weight is mg cos θ . ( N = mg 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 = (1 . 9 kg) (9 . 8 m / s 2 ) sin 32 ◦ 2 . 7 kg + 1 . 9 kg + 8 . 3 kg − (0 . 14) (1 . 9 kg) (9 . 8 m / s 2 ) cos 32 ◦ 2 . 7 kg + 1 . 9 kg + 8 . 3 kg + (8 . 3 kg − 2 . 7 kg) (9 . 8 m / s 2 ) 2 . 7 kg + 1 . 9 kg + 8 . 3 kg = 4 . 84778 m / s 2 . brown (twb493) – midterm 02 practice – turner – (57340) 2 002 (part 2 of 3) 10.0 points What is the tension in the cord connected to the 2 . 7 kg block? Correct answer: 39 . 549 N. Explanation: Using Eq. 1, we have T 1 = m 1 g + m 1 a = (2 . 7 kg) (9 . 8 m / s 2 + 4 . 84778 m / s 2 ) = 39 . 549 N . 003 (part 3 of 3) 10.0 points What is the tension in the cord connected to the 8 . 3 kg block? Correct answer: 41 . 1034 N....
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 Spring '08
 Turner
 Physics, Force, Friction, Mass, Sin, Correct Answer

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