Physics ch 6 answers

#30 F = 30NIf flat W = F N so W = N constant velocity  F f = F app = 30N 12 kg

#31so W = 44,100 N *To keep at constant height Lift Force = Weight = 44,100 N *To lift at 2 m/s 2 F = ma F acc = (4500 kg) (2 m/s 2 ) = 9000N F app = F g + F acc = 44,1000 N N = 53,100N 4500 kg

#32so W = 196 N So 196 needed to simply overcome g *To lift at 5 m/s 2 F = ma F acc = (20 kg) (5 m/s 2 ) = 100N Total F = F g + F acc = 196 N N = 296 N Since our max was 250 the bag will burst 20 kg

#33 F app = 40Nso W = 49 N F acc = (5 kg) (6 m/s 2 ) = 30 N F app = F f + F acc 40 N = F f + 30 N So F f = 10 N 5 kg

#34 so W = 2205 N If flat W = F N F f =441N to push at constant velocity F app = F f If pushing at 710N, 441N is for F f so F acc = 710 N – 441 N = 269 N = ma 269 N = (225 kg) a so a = m/s kg

#35v f = 0,v i = 14 m/s, d = 25m v f 2 = v i 2 + 2ad 0 = (14 m/s) 2 + 2(a)(25m) 0 = 196 m 2 /s m (a) -196 m 2 /s 2 = 50m (a) m/s 2 = a

#36a so W =F g = 836 N Moving up F app = 936N F app = F g + F acc 936 N = 836 N + F acc 100 N = F acc F acc = ma 100 N = (85.3 kg)(a) a = 1.17 m/s kg

#36b so W =F g = 836 N F app = 782 N F app = F g + F acc 782 N = 836 N + F acc -54 N = F acc F acc = ma -54 N = (85.3 kg)(a) a = m/s kg

#36 c & d c) Stopping because magnitude of acceleration is less d) Initial acc (neg) downwardscale < 836 N At Constant velocityscale = 836 N elevator goes to stop (acc is +)scale > 836 N

#37 a) W = mg = 490 N b) To start sled moving must overcome Static Friction F f static = 147N c) To keep constant velocity must overcome kinetic friction F f kinetic = 49 N d) To accelerate must overcome Kinetic Friction & have F acc so F app =F f + F acc 49 N + ma = 49 N + (50 kg)(3m/s 2 ) = 199 N 50 kg