phuchw6sol

phuchw6sol - Quach, Phuc Homework 6 Due: Sep 29 2006, 5:00...

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Unformatted text preview: Quach, Phuc Homework 6 Due: Sep 29 2006, 5:00 pm Inst: Mark Rupright 1 This print-out should have 17 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 Vector ~ A has components A x =- 5 , A y = 3 . 9 , A z = 2 . 1 , while vector ~ B has components B x = 6 . 2 , B y =- 5 . 3 , B z = 4 . 9 . What is the angle AB between these vec- tors? (Answer between 0 and 180 .) Correct answer: 130 . 62 . Explanation: Note: The magnitude of vector ~ X is k X k . Consider two formulae for the scalar prod- uct ~ A ~ B of two vectors: ~ A ~ B = A x B x + A y B y + A z B z (1) in terms of the two vectors components, and also ~ A ~ B = k ~ A k k ~ B k cos AB (2) in term of their magnitudes and the angle be- tween them. Given the data, we immediately calculate k ~ A k = q A 2 x + A 2 y + A 2 z = 6 . 67982 , (3) k ~ B k = q B 2 x + B 2 y + B 2 z = 9 . 51525 , (4) and using eq. (1), ~ A ~ B =- 41 . 38 . (5) Hence, according to eq. (2), cos AB = ~ A ~ B k ~ A k k ~ B k =- . 651037 (6) and therefore AB = arccos(- . 651037) = 130 . 62 . (7) Two vectors always lie in a plane. When these two vectors are plotted in this plane, we have A B 1 3 . 6 2 keywords: 002 (part 1 of 1) 10 points You drag a(n) 16 . 9 kg steamer trunk over a rough surface by a constant force of 79 . 2 N acting at an angle of 32 . 1 above the horizon- tal. You move the trunk over a distance of 13 . 7 m in a straight line, and the coefficient of kinetic friction is 0 . 122. The acceleration of gravity is 9 . 8 m / s 2 . 16 . 9 kg = 0 . 122 7 9 . 2 N 3 2 . 1 How much is the work done on the block by the net force? Correct answer: 712 . 688 J. Explanation: F mg n f k Work is W = ~ F ~s , where ~s is the distance traveled. In this problem ~s = 5 x is only in the x direction. W F = ~ F x ~s x = F s x cos = (79 . 2 N)(13 . 7 m)cos32 . 1 Quach, Phuc Homework 6 Due: Sep 29 2006, 5:00 pm Inst: Mark Rupright 2 = 919 . 161 J . To find the frictional force, F friction = N , we need to find N from vertical force balance. Note: N is in the same direction as the y component of F and opposite the force of gravity. Thus F sin + N = mg , so that N = mg- F sin . Thus the friction force is ~ F friction =- N x =- ( mg- F sin ) x, and the work done by friction is W = ~ F friction ~s =-| F f | | s | =- ( mg- F sin ) s x =- . 122 h (16 . 9 kg)(9 . 8 m / s 2 )- (79 . 2 N)sin32 . 1 i (13 . 7 m) =- 206 . 473 J . Since the normal force points in the y direc- tion and ~s is in the x direction W N = ~ N ~s = N y x = 0 ....
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This note was uploaded on 10/19/2008 for the course PHY 2048 taught by Professor Guzman during the Fall '08 term at FAU.

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phuchw6sol - Quach, Phuc Homework 6 Due: Sep 29 2006, 5:00...

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