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phuchw6sol - Quach Phuc Homework 6 Due 5:00 pm Inst Mark...

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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 θ 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 formulæ 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 θ 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 θ AB = ~ A · ~ B k ~ A k k ~ B k = - 0 . 651037 (6) and therefore θ AB = arccos( - 0 . 651037) = 130 . 62 . (7) Two vectors always lie in a plane. When these two vectors are plotted in this plane, we have A B 130 . 62 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 μ = 0 . 122 79 . 2 N 32 . 1 How much is the work done on the block by the net force? Correct answer: 712 . 688 J. Explanation: F θ m g n f k Work is W = ~ F · ~s , where ~s is the distance traveled. In this problem ~s = 5ˆ x is only in the ˆ x direction. W F = ~ F x · ~s x = F s x cos θ = (79 . 2 N) (13 . 7 m) cos 32 . 1
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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 = μ 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 θ + N = m g , so that N = m g - F sin θ . Thus the friction force is ~ F friction = - μ N ˆ x = - μ ( m g - F sin θ ) ˆ x , and the work done by friction is W μ = ~ F friction · ~s = -| F f | | s | = - μ ( m g - F sin θ ) s x = - 0 . 122 h (16 . 9 kg) (9 . 8 m / s 2 ) - (79 . 2 N) sin 32 . 1 i (13 . 7 m) = - 206 . 473 J . Since the normal force points in the ˆ y direc- tion and ~s is in the ˆ x direction W N = ~ N · ~s = N ˆ y · ˆ x = 0 . The net work done on the body is the alge- braic sum of the work done by the external force F and the work done by the frictional force W net = W F + W μ = (919 . 161 J) + ( - 206 . 473 J) = 712 . 688 J . keywords: 003 (part 1 of 1) 10 points A force ~ F = F x ˆ ı + F y ˆ acts on an object with F x = 4 N and F y = 2 N. The angle between ~ F and the displacement vector ~s is 33 , and 68 . 3 J of work is done by ~ F . Find the magnitude of ~s .
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