# Answer, Key Homework 7 David McIntyre Mar 25,

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1 Answer, Key Hoework 7 David McIntyre 453 Mar 5, 004 This print-out should have 4 questions. Multiple-choice questions ay continue on the next colun or page find all choices before aking your selection. The due tie is Central tie. Chapters 6 and 7 probles. 00 (part of ) 0 points A cheerleader lifts his 70.9 kg partner straight up off the ground a distance of before releasing her. The acceleration of gravity is 9.8 /s. If he does this 7 ties, how uch work has he done? Correct answer: 43.5 J. The work done in lifting the cheerleader once is g h (70.9 kg)(9.8 /s )(0.594 ) 4.73 J. The work required to lift her n 7 ties is n (7)(4.73 J) 43.5 J. 00 (part of 5) 0 points In the parallel spring syste, the springs are positioned so that the weight stretches each spring equally. The spring constant for the left-hand spring is 3.5 N/c and the spring constant for the righ-hand spring is 5.5 N/c. k 5.5 N/c,, and The springs stretch the sae aount x because of the way they were positioned. Then F k x and F k x, so the force equation for the suspended ass is Fup F down k x + k x x k + k 3.5 N/c N/c c. Diensional analysis for x: N N/c N c N c 003 (part of 5) 0 points In this sae parallel spring syste, what is the effective cobined spring constant k parallel of the two springs? Correct answer: 9 N/c. If the springs were one spring, that spring would react with a force F k x where F due the law of action and reaction, we have k parallel x 3.5 N/c 5.5 N/c How far down will the weight stretch the springs? Correct answer: c. Let : k 3.5 N/c, x c 9 N/c, which happens to be the su of the individual constants k parallel k + k 3.5 N/c N/c 9 N/c.

2 Answer, Key Hoework 7 David McIntyre 453 Mar 5, (part 3 of 5) 0 points Now consider the sae two springs in series. x total x +x, so k series x total, which is equivalent to 3.5 N/c 5.5 N/c hat distance will the spring of constant 3.5 N/c stretch? Correct answer:.574 c. In the series syste, the springs stretch a different aount, but each carries the full weight. k x and x k 3.5 N/c.574 c. 005 (part 4 of 5) 0 points In this sae series spring syste, what distance will the spring of constant 5.5 N/c stretch? Correct answer: 8 c. k x and k series x + x.574 c + 8 c.3889 N/c, or k series x + x x + x + k k 3.5 N/c N/c.3889 N/c. 007 (part of ) 5 points The pulley syste is in equilibriu, the spring constant k 7 N/c and the suspended ass 6 kg. The acceleration of gravity is 9.8 /s. x k 5.5 N/c 8 c. k 006 (part 5 of 5) 0 points In this sae series spring syste, what is the effective cobined spring constant k series of the two springs? Correct answer:.3889 N/c. How uch will the spring stretch? Correct answer: c. The existence of a spring in a string defines the tension in the string because the force exerted by a spring is F kx.

3 Answer, Key Hoework 7 David McIntyre 453 Mar 5, T T k T T k T 4 T T. Thus at the suspended ass, x g 3 k T + T g 3 T g 3 k x g (6 kg) ( 9.8 /s ) 3(7 N/c) c T 5 For the pulley suspended fro the spring, k x so that k x orking up fro the suspended weight, + T 5 so that T 5 k x For the lowest pulley, T 4 k x Thus at the weight, 008 (part of ) 5 points The pulley syste is in equilibriu, the spring constant k 8 N/c, the suspended weight 39 N, and the suspended weight N. k x + T 4 + T 5 + k x + k x k x k x k 39 N N 8 N/c c How uch will the spring stretch? Correct answer: c. 009 (part of ) 5 points A crate is pulled by a force (parallel to the incline) up a rough incline. The crate has an initial speed shown in the figure below. The crate is pulled a distance of 6.3 on the incline by a 50 N force. The acceleration of gravity is 9.8 /s.

4 Answer, Key Hoework 7 David McIntyre 453 Mar 5, /s 3 kg µ N 9 The work due to the applied force is appl F d (50 N) (6.3 ) 945 J, and the work due to gravity is a) hat is the change in kinetic energy of the crate? Correct answer: J. Let : F 50 N, d 6.3, θ 9, 3 kg, g 9.8 /s, µ 0.38, and v.56 /s. N µ N g The work-energy theore with nonconservative forces reads fric + appl + gravity K To find the work done by friction we need the noral force on the block fro Newton s law Fy N g cos θ 0 Thus v N g cos θ. fric µ g d cos θ (0.38) (3 kg) (9.8 /s ) (6.3 ) cos J. θ F so that grav g d sin θ (3 kg) (9.8 /s ) (6.3 ) sin J, K fric + appl + grav ( 30.5 J) + (945 J) + ( J) J. 00 (part of ) 5 points b) hat is the speed of the crate after it is pulled the 6.3? Correct answer: /s. Since (v f v i ) K v f v i K K v f + v i (35.63 J) + (.56 /s) 3 kg /s. 0 (part of 4) 0 points You drag a suitcase of ass 8. kg with a force of F at an angle 3.6 with respect to

5 Answer, Key Hoework 7 David McIntyre 453 Mar 5, the horizontal along a surface with kinetic coefficient of friction The acceleration of gravity is 9.8 /s. If the suitcase is oving with constant velocity.99 /s, what is F? Correct answer: N. F 04 (part 4 of 4) 0 points If you are accelerating the suitcase with acceleration /s what is F? Correct answer: N. If the suitcase is accelerating with an acceleration a, the force F is F cos θ µ ( g F sin θ) a g If the suitcase is oving at a constant velocity, F cos θ µ N µ ( g F sin θ) F N µ g cos θ + µ sin θ 0.36 (8. kg) (9.8 /s ) cos sin N. 0 (part of 4) 0 points hat is the noral force on the suitcase? Correct answer: N. You are helping support the weight, so the noral force is N g F sin θ (8. kg) (9.8 /s ) (7.468 N) sin N. 03 (part 3 of 4) 0 points If you pull the suitcase 98.9, what work have you done? Correct answer: 33.8 J. w F d cos θ (7.468 N) (98.9 ) cos J. F a + µ g cos θ + µ sin θ (8. kg) (0.947 /s ) cos sin (8. kg) (9.8 /s ) cos sin N. 05 (part of 4) 3 points A 5.7 kg box initially at rest is pushed.36 along a rough, horizontal floor with a constant applied horizontal force of N. The acceleration of gravity is 9.8 /s. If the coefficient of friction between box and floor is 0.468, find the work done by the applied force. Correct answer: J. The work done by the applied force is given by F F s (6.303 N)(.36 ) J 06 (part of 4) 3 points Find the work done by the friction. Correct answer: J. The work done by the friction is given by f f s µ g s f 0.468(5.7 kg)(9.8 /s )(.36 ) (part 3 of 4) points Find the change in kinetic energy of the box. Correct answer: J.

6 Answer, Key Hoework 7 David McIntyre 453 Mar 5, The net work done is given by Then, using Kepler s third law, these four quantities are related by net F + f J+( J) J T c T e a3 c a 3, e 08 (part 4 of 4) points Find the the final speed of the box. Correct answer:.3493 /s. The change in kinetic energy is equal to the net work done, so K net v f The initial velocity is zero, so net v f (46.56 J) 5.7 kg.3493 /s v i 09 (part of 3) 4 points The distance of closest approach of Halley s coet to the sun is k ( 0.57 AU). The period of the coet is 75.6 years. The radius of the earth s orbit around the sun is k ( AU) (assue the earth s orbit is circular). Find the length of the sei-ajor axis of the coet s orbit. Correct answer: AU. Basic Concepts: Kepler s Third Law. Solution: to the sei-ajor axis, a, by Kepler s third law, Let and ( 4 π T ) a 3. G M sun T e Period of Earth s orbit, a e Sei-ajor axis of Earth s orbit, T c Period of coet s orbit, a c Sei-ajor axis of coet s orbit. since the factor of proportionality cancels out. Hence, solving for a c, one gets ( T ) /3 a c a c e Te ( 75.6 years ( AU ) years AU. ) /3 00 (part of 3) 3 points Denote the length of the sei-ajor axis of the orbit by a, and the distance of the closest approach to the sun by d. The axiu distance between the Halley s coet and the sun is given by. a d correct. (a d) 3. a d 4. (a + d) 5. a + d 6. a + d a (Refer to the above figure.) d s 0 (part 3 of 3) 3 points hat is the axiu distance between the x

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