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Unformatted text preview: homework 05 – VENNES, ROSS – Due: Oct 5 2007, 1:00 am 1 Question 1, chap 7, sect 1. part 1 of 2 10 points A 15 . 5 kg block is dragged over a rough, horizontal surface by a constant force of 177 N acting at an angle of 32 . 8 ◦ above the horizon tal. The block is displaced 46 . 8 m, and the coefficient of kinetic friction is 0 . 162. The acceleration of gravity is 9 . 8 m / s 2 . 15 . 5 kg μ = 0 . 162 1 7 7 N 3 2 . 8 ◦ Find the work done by the 177 N force. Correct answer: 6962 . 92 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 = (177 N) (46 . 8 m) cos32 . 8 ◦ = 6962 . 92 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: 424 . 703 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 )ˆ 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 = . 162[(15 . 5 kg) (9 . 8 m / s 2 ) (177 N) sin32 . 8 ◦ ] (46 . 8 m) = 424 . 703 J . Question 3, chap 7, sect 1. part 1 of 1 10 points If 3.1 J of work is done in raising a 177 g apple, how far is it lifted? Correct answer: 1 . 78533 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 = 3 . 1 J m = 177 g g = 9 . 81 m / s 2 Solution: The force and displacement are parallel, so the distance is given by homework 05 – VENNES, ROSS – Due: Oct 5 2007, 1:00 am 2 d = W applied mg = 3 . 1 J (177 g)(9 . 81 m / s 2 ) · 1000 g 1 kg = 1 . 78533 m Question 4, chap 7, sect 1. part 1 of 1 10 points A tugboat exerts a constant force of 5810 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 . 95 km? Correct answer: 28 . 7595 MJ (tolerance ± 1 %). Explanation: Given : F net = 5810 N and d = 4 . 95 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 = (5810 N) (4 . 95 km) parenleftbigg 1000 m 1 km parenrightbigg · parenleftbigg 1 MJ 10 6 J parenrightbigg = 28 . 7595 MJ , since 1 J = 1 N · m. Question 5, chap 7, sect 1....
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This homework help was uploaded on 04/18/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Force, Work

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