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Unformatted text preview: Morby, Grant – Homework 11 – Due: Feb 15 2006, noon – Inst: Drummond 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The force required to stretch a spring varies directly with the amount the spring is stretched. A force of 11 pounds is needed to stretch a spring 66 inches, as shown in the righthand figure below. 57 00 66 00 W 11 lbs How much force is required to stretch the spring 57 inches? Correct answer: 9 . 5 lbs. Explanation: Given : F 2 = 11 lbs , x 2 = 66 inches , and x 1 = 57 inches . F ∝ x , so F = k x = ⇒ k = F x . Thus F 1 x 1 = F 2 x 2 = F 2 x 1 x 2 = (11 lbs)(57 inches) 66 inches = 9 . 5 lbs . 002 (part 1 of 3) 10 points The suspended 2 . 4 kg mass on the right is moving up, the 2 . 2 kg mass slides down the ramp, and the suspended 7 . 6 kg mass on the left is moving down. The coefficient of friction between the block and the ramp is 0 . 13 . The acceleration of gravity is 9 . 8 m / s 2 . The pulleys are massless and frictionless. 2 . 2 k g μ = . 1 3 32 ◦ 7 . 6 kg 2 . 4 kg What is the acceleration of the three block system? Correct answer: 4 . 9187 m / s 2 . Explanation: Let : m 1 = 2 . 4 kg , m 2 = 2 . 2 kg , m 3 = 7 . 6 kg , and θ = 32 ◦ . Basic Concept: F net = ma 6 = 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 Morby, Grant – Homework 11 – Due: Feb 15 2006, noon – Inst: Drummond 2 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 = μm 2 g cos θ act up the table F net 2 = m 2 a (2) = T 3 + m 2 g sin θ T 1 μm 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 μm 2 g cos θ m 1 g . Solving for a , we have a = [ m 2 sin θ μm 2 cos θ + ( m 3 m 1 )] g m 1 + m 2 + m 3 = (2 . 2 kg)(9 . 8 m / s 2 ) sin32 ◦ 2 . 4 kg + 2 . 2 kg + 7 . 6 kg (0 . 13)(2 . 2 kg)(9 . 8 m / s 2 ) cos32 ◦ 2 . 4 kg + 2 . 2 kg + 7 . 6 kg + (7 . 6 kg 2 . 4 kg)(9 . 8 m / s 2 ) 2 . 4 kg + 2 . 2 kg + 7 . 6 kg = 4 . 9187 m / s 2 ....
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 Spring '09
 KLEINMAN
 Physics, Force, Friction, Work, kg, noon – Inst

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