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physics hw 5

# physics hw 5 - homework 05 HILL WILLIAM Due Oct 5 2007 1:00...

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homework 05 – HILL, WILLIAM – Due: Oct 5 2007, 1:00 am 1 Question 1, chap 7, sect 1. part 1 of 2 10 points A 16 . 7 kg block is dragged over a rough, horizontal surface by a constant force of 114 N acting at an angle of 33 . 6 above the horizon- tal. The block is displaced 50 . 5 m, and the coefficient of kinetic friction is 0 . 184. The acceleration of gravity is 9 . 8 m / s 2 . 16 . 7 kg μ = 0 . 184 114 N 33 . 6 Find the work done by the 114 N force. Correct answer: 4795 . 13 J (tolerance ± 1 %). Explanation: Consider the force diagram F θ m g n f k Work is W = vector F · vectors , where vectors is the distance traveled. In this problem vectors = 5ˆ ı is only in the x direction. W F = F x s x = F s x cos θ = (114 N) (50 . 5 m) cos 33 . 6 = 4795 . 13 J , . Question 2, chap 7, sect 1. part 2 of 2 10 points Find the magnitude of the work done by the force of friction. Correct answer: 934 . 528 J (tolerance ± 1 %). Explanation: To find the frictional force, F friction = μ N , we need to find N from vertical force balance. Note that 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 vector F friction = - μ N ˆ ı = - μ ( m g - F sin θ ı . The work done by friction is then W μ = vector F friction · vectors = -| f μ | | s | = - μ ( m g - f μ sin θ ) s x = - 0 . 184 [(16 . 7 kg) (9 . 8 m / s 2 ) - (114 N) sin 33 . 6 ] (50 . 5 m) = - 934 . 528 J . Question 3, chap 7, sect 1. part 1 of 1 10 points If 2.2 J of work is done in raising a 180 g apple, how far is it lifted? Correct answer: 1 . 24844 m (tolerance ± 1 %). Explanation: Basic Concepts: W applied = F applied d cos θ = mgd since θ = 0 cos θ = 1. 1 J = 1 N · m F = mg 1 N = 1 kg · m / s 2 Given: W applied = 2 . 2 J m = 180 g g = 9 . 81 m / s 2 Solution: The force and displacement are parallel, so the distance is given by

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homework 05 – HILL, WILLIAM – Due: Oct 5 2007, 1:00 am 2 d = W applied mg = 2 . 2 J (180 g)(9 . 79 m / s 2 ) · 1000 g 1 kg = 1 . 24844 m Question 4, chap 7, sect 1. part 1 of 1 10 points A tugboat exerts a constant force of 2900 N on a ship moving at constant speed through a harbor. How much work does the tugboat do on the ship if each moves a distance of 4 . 42 km? Correct answer: 12 . 818 MJ (tolerance ± 1 %). Explanation: Given : F net = 2900 N and d = 4 . 42 km . The net force F net = Σ F x = F . The force and the displacement are parallel, so the net work is W net = F net d = (2900 N) (4 . 42 km) parenleftbigg 1000 m 1 km parenrightbigg · parenleftbigg 1 MJ 10 6 J parenrightbigg = 12 . 818 MJ , since 1 J = 1 N · m. Question 5, chap 7, sect 1. part 1 of 3 10 points A 2 . 1 kg block is pushed 1 . 52 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of 69 . 2 with the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 1 kg F 69 . 2 If the coefficient of kinetic friction between the block and wall is 0 . 673, find the work done by F . Correct answer: 42 . 0253 J (tolerance ± 1 %).
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physics hw 5 - homework 05 HILL WILLIAM Due Oct 5 2007 1:00...

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