PHY
HW 11

# HW 11 - Morby Grant Homework 11 Due noon Inst Drummond This...

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Morby, Grant – Homework 11 – Due: Feb 15 2006, noon – Inst: Drummond 1 This print-out should have 11 questions. Multiple-choice 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 right-hand 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 kg μ = 0 . 13 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 = m a 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

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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 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 . 2 kg) (9 . 8 m / s 2 ) sin 32 2 . 4 kg + 2 . 2 kg + 7 . 6 kg - (0 . 13) (2 . 2 kg) (9 . 8 m / s 2 ) cos 32 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 . 003 (part 2 of 3) 10 points What is the tension in the cord connected to the 2 . 4 kg block? Correct answer: 35 . 3249 N. Explanation: Using Eq. 1, we have T 1 = m 1 g + m 1 a = (2 . 4 kg) (9 . 8 m / s 2 + 4 . 9187 m / s 2 ) = 35 . 3249 N .
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