= tan-1 (Ry/Rx)  = tan-1 (12.76/5.85)  = tan-1 (2.181)  = south of east  PTS: 1 DIF: 4 REF: p. 123 Bloom’s Level NAT: B.4 STA: P3.4e 23. ANS: Fnet = F = F = ma kFN kmg = (F ma) / (mg) k k = [68 N (53 kg)(0.10 m/s2)] / [(53 kg)(9.8 m/s2)] = (62.7 N) / (519.4 N) k = 0.12 k PTS: 1 DIF: 5 REF: pp. 126-127 Bloom’s Level NAT: B.4 STA: P3.2C | P3.4e 24. ANS: Fnet = F - = F mg = ma kFN k = (F ma) / mg k = [80.0 N (250 kg)(0.08 m/s2)] / [(250 kg)(9.8 m/s2)] k = (80.0 N 20 N) / (2450 N) k = 0.024 k PTS: 1 DIF: 5 REF: pp. 126-127 Bloom’s Level NAT: B.4 STA: P3.2C | P3.4e
 = tan-1 (Ry/Rx)  = tan-1 (12.76/5.85)  = tan-1 (2.181)  = south of east PTS: 1 DIF: Bloom’s Level 4 REF: p. 123 NAT: B.4 STA: P3.4e 23. ANS: Fnet = F kFN = F kmg = ma k = (F ma) / (mg) k = [68 N (53 kg)(0.10 m/s2)] / [(53 kg)(9.8 m/s2)] k = (62.7 N) / (519.4 N) k = 0.12 PTS: 1 DIF: Bloom’s Level 5 REF: pp. 126-127 NAT: B.4 STA: P3.2C | P3.4e 24. ANS: Fnet = F - kFN = F k mg = ma k = (F ma) / mg k = [80.0 N (250 kg)(0.08 m/s2)] / [(250 kg)(9.8 m/s2)] k = (80.0 N 20 N) / (2450 N) k = 0.024 PTS: 1 DIF: Bloom’s Level 5 REF: pp. 126-127 NAT: B.4 STA: P3.2C | P3.4e
5.1-5.2_quiz_review.docx