This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 = 0 . 184 1 1 4 N 3 3 . 6 Find the work done by the 114 N force. Correct answer: 4795 . 13 J (tolerance 1 %). Explanation: Consider the force diagram F m g n f k Work is W = vector F vectors , where vectors is the distance traveled. In this problem vectors = 5 is only in the x direction. W F = F x s x = F s x cos = (114 N) (50 . 5 m) cos33 . 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 = 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 + N = m g so that N = m g F sin . Thus the friction force is vector F friction = N = ( m g F sin ) . The work done by friction is then W = vector F friction vectors = f  s  = ( m g f sin ) s x = . 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 = mgd since = 0 cos = 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 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 = 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....
View
Full
Document
This note was uploaded on 04/24/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Force, Work

Click to edit the document details