## Answers

Tg a) m= 6kg t = 1.2 +3 ? + sigt ĵ – 3.3 ty Ê o = dr = 3.6 t ì + 5.9 9 – 13 12 t3 k a = 7.2 t i to-39.6X (2 Ê a= 7o2ti - 39.6 € E a lat tosa 7.2 (5) ? - 39.6 (5) ť To To = -954 m/8² F =ma = 6x -954 = -572uN -5724 N 8= 227 m. g=lomis u=32mls FS < AS and As N=ng fo= Mg N = Meng - ez u? = 2a8 (0)²-(32)2 = 2x ax 227 - (32) - 2.256 2x227 - wa= Meng a= lsg & le- Scanned a

2.256 Mp= 0.23 J su c) m= 6.3 x 103 kg, u= 20ml, v= 16m18 Conceptually since initial its net projectile returns to its position again , therefore displacement is zero o work done by = W = F. (0) wt g projectile [w=of my But if their is no friction & frictio air then - work done by wt g am(4²3) projectile = I x 6.3 x103 (2016) I w = 45.36 X (62 J my Scanned with CamScanner

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