Answers to selected problems from Essential Physics, Chapter 3
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1 Answers to selected problems from Essentil Physics, Chpter 3 1. FBD 1 is the correct free-body dirm in ll five cses. As fr s forces re concerned, t rest nd constnt velocity situtions re equivlent. 3. () FBD 1 (b) FBD 3 (c) FBD 2 5. The penny nd the fether fll t the sme rte. In the bsence of ir resistnce, the only force tht mtters is the force of rvity. Thus, the ccelertion of both objects is the ccelertion due to rvity, which is the sme for ny object. 7. () The net force on ech of the blocks is zero. (b) Both block 2 nd block 3 experience three forces. 1 they both experience force of rvity, pplied by the Erth, which is the sme for both blocks. 2 They both experience downwrd norml force, pplied by the block bove it. The downwrd norml force pplied by block 2 on block 3 is lrer thn tht pplied by block 1 on block 2, however, becuse this norml force is equl to the weiht of the blocks sittin on top of the block feelin the norml force. 3 They both experience n upwrd norml force, pplied by the object immeditely below them. The norml force pplied by the floor on block 3, however, hs mnitude equl to the weiht of three blocks, while the upwrd norml force pplied by block 3 on block 2 hs mnitude equl to the weiht of two blocks. 9. () Any constnt-velocity motion, such s trvelin t constnt velocity lon striht hihwy in cr, hockey puck slidin cross frictionless ice surfce, or spce probe driftin throuh spce, trillions of kilometers from nythin else. (b) Not possible. (c) This is possible s lon s there re 2 or more forces ctin on the object, nd ll the forces cncel. An exmple could be you sittin on chir. 11. Yuri could throw the wrench directly wy from the spce sttion. To do this, he would exert force on the wrench in the direction he throws the wrench. By Newton s third lw, the wrench would exert n equl-nd-opposite force bck on Yuri. Once Yuri let o of the wrench, he would drift t constnt velocity bck towrd the spce sttion Essentil Physics, Answers to selected Chpter 3 Problems Pe 1
2 17. (f) We just need two different free-body dirms. In both () nd (e), the bll s ccelertion is directed up, so the upwrd norml force pplied by your hnd is lrer thn the downwrd force of rvity. In (b), the only thin ctin on the bll is the force of rvity, nd whether the bll is movin up, down, t instntneously t rest is irrelevnt Σ F = m x N = m (3.0 m/s ) 2 (e) 2.0 k 23. Σ F = ( M + m) = 0 + F F = 0 N,from tble G,totl Σ F = ( M) = 0 + F F F = 0 N,from tble G2 N1 (e) 10 N, directed down (f) 50 N, directed up Essentil Physics, Answers to selected Chpter 3 Problems Pe 2
3 25. Σ F = ( M + m) = 0 + F F 10 N = 0 N,from tble G,totl Σ F = ( M) = 0 + F F F = 0 N,from tble G2 N1 (e) 20 N, directed down (f) 60 N, directed up 27. Σ F = ( M + m) + F F = ( M + m) + FT1 ( M + m) = ( M + m) T1 G,totl 5 FT1 = ( M + m ) Σ F = ( M) + FT2 FG 2 = M + FT2 M = M 5 FT2 = M (e) 15 N (f) 10 N 29. Essentil Physics, Answers to selected Chpter 3 Problems Pe 3
4 Σ F = (2 m+ m) + 30 N = 3m 10 N = m Σ F = ( m) 10 N + FT 2 = m m F = 10 N T 2 (e) () 20 m/s est (b) 30 m/s est (c) () N (b) 3800 N N 37. Up until t = 5 s, there is no net force pplied to the object. For 1 s, between t = 5 s nd t = 6 s, constnt force directed to the riht is pplied. After t = 6 s, no net force is pplied N 1. () The fifth floor. (b) 20% of ( little under 2.0 m/s 2 ) 3. () 0.00 m/s 2 (b) 350 m/s (c) 200 m/s 5. The mximum ccelertion, of 3.6 m/s 2, is when ll three forces re in the sme direction. The minimum ccelertion is zero, becuse the forces cn be rrned so tht they cncel out N 9. () 8T/5 (b) 2T 51. () equl in the two cses (b) F/5 in both cses (c) in cse 2 (i) F/2 (ii) F/5 53. () 1.0 N (b) 2.0 N (c) 5.0 N 55. () No, it s the sme free-body dirm. (b) No, it s lso the sme. In ech cse, the bll is bein cted upon only by rvity. 57. () (100 m/s 2 ) m (b) 2 m/s, in the direction the puck ws trvelin initilly m 59. () FMm = F M + m M (b) FMm = F M + m (c) M m = Essentil Physics, Answers to selected Chpter 3 Problems Pe
5 61. () The elevtor is t rest for the first 2 seconds, nd then ccelertes down for 2 seconds. This is followed by -second period of constnt velocity, directed down, followed by 2-second period of upwrd ccelertion, while the elevtor is movin down but in the process of comin to rest. The elevtor remins t rest fter tht. (b) 2.0 m/s 2 (c) The elevtor moved down totl distnce of 2 m. Essentil Physics, Answers to selected Chpter 3 Problems Pe 5
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