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06homework05

# 06homework05 - Husain Zeena – Homework 5 – Due 4:00 am...

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Unformatted text preview: Husain, Zeena – Homework 5 – Due: Sep 30 2003, 4:00 am – Inst: H L Berk 1 This print-outshouldhave24questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. 001 (part 1 of 2) 10 points Given: g = 9 . 8 m / s 2 . A block starts from rest at a height of 3 . 2 m on a fixed inclined plane. 3 . 7 k g μ = . 2 6 28 ◦ What is the speed of the block at the bot- tom of the ramp? Hint: This problem requires a train of logic. (1) Analyze force diagram, (2) use Newton’s Laws, and (3) solve the equations of motion. Correct answer: 5 . 66133 m / s. Explanation: Let : h = 3 . 2 m , m = 3 . 7 kg , μ = 0 . 26 , θ = 28 ◦ , and v f = final speed . F f N mg 28 ◦ d h 28 ◦ The normal force to the inclined plane is N = mg cos θ . The sum of the forces par- allel to the inclined plane is F net = ma = mg sin θ- μmg cos θ a = g sin θ- μg cos θ Since v 2 f = v 2 + 2 ax = 2 ad (1) along the plane and the distance moved along the plane is d = h sin θ (2) therefore v f = r 2 ah sin θ = r 2 g h (sin θ- μ cos θ ) sin θ = p 2 g h (1- μ cot θ ) (3) = q 2(9 . 8 m / s 2 )(3 . 2 m)[1- (0 . 26)cot28 ◦ ] = 5 . 66133 m / s . 002 (part 2 of 2) 10 points If the block continues to slide on the ground with the same coefficient of friction, how far will the block slide on the ground until coming to rest? Correct answer: 6 . 28937 m. Explanation: F f N mg Here the normal force to the x axis is N = mg . The sum of the forces parallel to the x axis is F net = ma = μmg Husain, Zeena – Homework 5 – Due: Sep 30 2003, 4:00 am – Inst: H L Berk 2 Since v 2 f = v 2 + 2 ax v 2 =- 2 ax =- 2 μg x then x =- v 2 2 g μ | x | = (5 . 66133 m / s) 2 2(9 . 8 m / s 2 )(0 . 26) = 6 . 28937 m . 003 (part 1 of 1) 10 points Given: There is friction between the block and the table. g = 9 . 8 m / s 2 . The suspended 2 kg mass on the left is moving up, the 3 kg mass slides to the right on the table, and the suspended mass 4 kg on the right is moving down. 2 kg 3 kg 4 kg μ = 0 . 18 What is the magnitude of the acceleration of the system? Correct answer: 1 . 58978 m / s 2 . Explanation: m 1 m 2 m 3 μ a Let : m 1 = 2 kg , m 2 = 3 kg , m 3 = 4 kg , and μ = 0 . 18 . Basic Concepts: The acceleration a of each mass is the same, but the tensions in the two strings will be different. F net = ma 6 = 0 Solution: Let T 1 be the tension in the left string and T 2 be the tension in the right string. Consider the free body diagrams for each mass T 1 m 1 g a T 2 m 3 g a T 1 T 2 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, so F net 1 = m 1 a = T 1- m 1 g . (1) For the mass on the table, a is directed to the right, T 2 acts to the right, T 1 acts to the left, and the motion is to the right so that the frictional force μm 2 g acts to the left and F net 2 = m 2 a = T 2- T 1- μm 2 g . (2) For the mass...
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06homework05 - Husain Zeena – Homework 5 – Due 4:00 am...

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