CH 4 Mixed Review 46,50,53,55,56,59,61,&63

1. CH 4 Mixed Review 46,50,53,55,56,59,61,&63 CH 4 Mixed Review 46,50,53,55,56,59,61,&63

2. FN FA W F|| FA cos30 = 24.5 FA cos30 = 46 m = 5 W = 49 W= 42.4 W|| = 24.5 Equations from Diagram: W|| = F|| FN = F + W F|| F W|| W  30° Now that you know FA = 28.3 , you can use sin30 to get F FN = F + W FN = 14.2 + 42.4 Same as W|| FN = 56.6 N FA = 28.3 N

3. 50 T - W = FNET T W T - 5(9.8) = 5(3) T = 64 N

4. 53 FN W FN = W Ff = FNET Ff Find Ff: same as FNET Find a: Find d: FNET = m a vi = 7 vf = 0 d = ? a = vf2 = vi2 + 2ad Ff =  FN 32.3 = 65.8a 02 = 72 + 2(.491)d Ff = .05(645) a = .491 d = 49.9 m .491 Ff = 32.3

5. FN Ff W 55 m = 3 W=29.4 W=25.5 W||= 14.7 W|| - Ff = FNET FN = W W|| W  30° C. Already found this ! 9.36 N A. vi =0 d = 2 t = 1.5 a = ? B. W|| - Ff = FNET 14.7 – Ff = 3(1.78) Ff = 9.36 N d = vit + ½ at2 2 = ½ a(1.5)2 D. vf2 = vi2 + 2ad a = 1.78 m/s2 vf2 = 2(1.78)2 vf = 2.67 m/s Ff =  FN 9.36 = (25.5)  = .367

6. 56 Ff W W – Ff = FNET Find FNET : Find a : Find vf : W – Ff = FNET FNET = ma vi = 0 vf = ? d = 25 a = vf2 = vi2 + 2ad 75(9.8) – 95 = FNET vf2 = 2(8.53)(25) 640 = 75a FNET = 640 vf = 20.7 m/s 8.53 a = 8.53

7. FN Ff 35° W FA 59 m = 3 W = 29.4 W= 24.1 W|| = 16.9 W|| = Ff W + FA = FN W|| W  Find FN: Find FA : W + FA = FN Ff =  FN 24.1 + FA = 56.3 16.9 = .300FN FA = 32.2 N FN = 56.3

8. FN FA ° W W|| W sin = Same as FA 300 12200 sin = 61 W|| = FA W = FN W|| W   = 1.4°

9. B. Resultant force is NET force 4 For m3 For m2 For m1 3 FNET = ma FNET = ma FNET = ma 180 2 = 4(20) = 80 N = 3(20) = 60 N = 2(20) = 40 N A. Whole: FN FN FNET = ma FA Ffrom m1 = 7(20) = 140 N W W D. Between m2 and m3 : 63 C. Between m1 and m2: FNET = ma 180 = 9a a = 20 m/s2 80 N…..it is the NET force on m3 For ALL blocks

10. The End