Problem 4B - Newton's Second Law

Givens7. F net = −65.0 Nm = 0.145 kgSolutionsF −65.0 Na net = ⎯ net= ⎯ = −448 m/s 2m 0.145 kg8. m = 214 kgF net = F buoyant − mg = 790 N − (214 kg)(9.81 m/s 2 )F buoyant = 790 Ng = 9.81 m/s 2 F net = 790 N − 2.10 × 10 3 N = −1310 NF −1310 Na net = ⎯ net= ⎯ = −6.12 m/s 2m 214 kg9. m = 0.080 kgF net = m a net = m g(sin q)q = 37.0°a net = g(sin q) = (9.81 m/s 2 )(sin 37.0°) = 5.90 m/s 2g = 9.81 m/s 2a net = 5.90 m/s 2 down the incline (37.0° below horizontal)10. m = 0.080 kgF upward = 1.40 Na downward = 5.90 m/s 2F net = F upward − m a downward = 1.40 N − (0.080 kg)(5.90 m/s 2 ) = 1.40 N − 0.47 N = 0.93 NF net = 0.93 N up the incline (37.0° above the horizontal)a net = F net 0.93N⎯⎯ = ⎯⎯ = 12 m/s 2m 0 .080kgAdditional Practice 4C1. F downward = 4.26 × 10 7 Nm k = 0.25F net = F downward − F k = 0F k = m k F n = F downwardF 4.26 × 10 7 NF n = ⎯ downward= ⎯⎯ = 1.7 × 10 8 Nmk 0.25V2. F n = 1.7 × 10 8 NF n = mg (cos q)q = 10.0°F 1.7 × 10 8 Ng = 9.81 m/s 2 m = ⎯ n= ⎯⎯⎯ = 1.8 × 10 7 kgg (cos q) (9.81 m/s 2 )(cos 10.0°)3. F s,max = 2400 Nms = 0.20q = 30.0°g = 9.81 m/s 24. m = 60.0 kga = 3.70 m/s 2ms = 0.455g = 9.81 m/s 2F s,max = m s F nF 2400 NF n = ⎯ s,max= ⎯ = 1.2 × 10 4 Nms 0.20F n = 1.2 × 10 4 N perpendicular to and away from the inclineF n = mg(cos q)F 1.2 × 10 4 nNm = ⎯ = ⎯⎯⎯ = 1400 kgg (cos q) (9.81 m/s 2 )(cos 30.0°)For the passenger to remain standing without sliding,F s,max ≥ F = maF s,max = m s F n = m s mgm s mg ≥ mam s g ≥ a(0.455)(9.81 m/s 2 ) = 4.46 m/s 2 > 3.70 m/s 2Copyright © by Holt, Rinehart and Winston. All rights reserved.The passenger will be able to stand without sliding.V Ch. 4–4Holt Physics Solution Manual