Answer, Key Homework 7 David McIntyre Mar 25,


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1 Answer, Key Hoework 7 David McIntyre 453 Mar 5, 004 This printout should have 4 questions. Multiplechoice 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 lefthand spring is 3.5 N/c and the spring constant for the righhand 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 workenergy 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 seiajor axis of the coet s orbit. Correct answer: AU. Basic Concepts: Kepler s Third Law. Solution: to the seiajor 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 Seiajor axis of Earth s orbit, T c Period of coet s orbit, a c Seiajor 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 seiajor 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
7 Answer, Key Hoework 7 David McIntyre 453 Mar 5, Halley s coet and the sun? Correct answer: AU. x (7.879 AU ) 0.57 AU AU. 0 (part of ) 0 points Given: G N /kg. The planet Mars requires.4 years to orbit the sun, which has a ass of 0 30 kg, in an alost circular trajectory. Calculate the radius of the orbit of Mars as it circles the sun. Correct answer: and Thus, F r G M r a r ω r, ω π T π ( s) ( G M r rad/s. ω ) 3 [ 0 30 kg ( rad/s) ] 3 A synchronous satellite, which always reains above the sae point on a planet s equator, is put in orbit about Jupiter to study the Great Red Spot. Jupiter rotates once every 9.9 h, has a ass of kg and a radius of Given that G N /kg, calculate how far above Jupiter s surface the satellite ust be. Correct answer: Basic Concepts: Solution: According to Kepler s third law: T 4 π G M r3 where r is the radius of the satellite s orbit. Thus, solving for r: ( G M T ) 3 r 4 π ( N /kg M (35640 s) 4 ( ) Now, the altitude h of the satellite (easured fro the surface of Jupiter) is h r R ) 3 ( N /kg ) (part of ) 0 points Calculate the orbital speed of Mars as it circles the sun. Correct answer: 9 /s. v ω r ( rad/s) ( ) 9 /s. 04 (part of ) 0 points
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