until an unbalanced force acts on it!!

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Presentation transcript:

until an unbalanced force acts on it!! 1) It stays there forever until an unbalanced force acts on it!! Weight (Fg) down Table (FN) up FN NO Fg 2) It accelerates in the direction of the push Weight (Fg) down FN f Table (FN) up opposite the direction it moves Friction (f) Fg Yes Friction slows the ball until is stops

ΣF = 0 3) means the sum of the forces equals zero (the net force is zero) NO It means the object cannot be accelerating (it could be moving, but if it is moving, it’s moving with constant velocity) ΣF = ma 4) acceleration is directly proportional to the net applied force acceleration direction is in the direction of the net applied force 5) Given: Fg = 485 N g = 9.8 m/s2 Fg = mg m = 49.5 kg 485 = m(9.8) m =49,500 g Fg = (49,500)(980) Fg = 4.85 x 107 dynes

Fengine = 4800 Newtons Given: m = 1200 kg g = 9.8 m/s2 Fg = 11,760 N 6) Given: m = 1200 kg g = 9.8 m/s2 Fg = 11,760 N a = 4.0 m/s2 FN Fengine F = Fengine F = ma ma = Fengine (1200) (4) = Fengine Fg a Fengine = 4800 Newtons

a = 3.75 m/s2 Given: m = 400 kg g = 9.8 m/s2 Fg = 3920 N 7) Given: m = 400 kg g = 9.8 m/s2 Fg = 3920 N Fengine = 2250 N f = 750 N FN Fengine f F = Fengine - f F = ma ma = Fengine - f (400) (a) = 2250 - 750 Fg a a = 3.75 m/s2

Given: m = 6.0 kg g = 9.8 m/s2 Fg = 58.8 N (a) (b) FT FT a a = 0 Fg Fg 8) (a) (b) Fg = 58.8 N FT FT a a = 0 Fg Fg (c) (d) FT FT a a Fg Fg

FT= 58.8 N FT = 24.6 N FT = 81.0 N FT = 0.0 N F = FT - Fg F = ma ma = FT - Fg (6) (0) = FT - 58.8 F = Fg- FT F = ma ma = Fg - FT (6) (5.7) = 58.8 - FT (a) (b) FT= 58.8 N FT = 24.6 N F = FT - Fg F = ma ma = FT - Fg (6) (3.7) = FT - 58.8 F = Fg- FT F = ma ma = Fg - FT (6) (9.8) = 58.8 - FT (d) (c) FT = 81.0 N FT = 0.0 N

ΣF = 82.65 N Fg = mg 7.35 = m(9.8) a = 110 m/s2 mAgena = 3292 kg Given: Fupward = 90 N Fg = 7.35 N 9) Fupward a Fg (a) Fg = mg 7.35 = m(9.8) m = 0.750 kg (750 grams) (b) F = Fupward - Fg F = 90 – 7.35 ΣF = 82.65 N F = ma 82.65 = (0.75) (a) (c) a = 110 m/s2 F = Faverage F = ma mAgena = 3292 kg 890 = m (0.133) Given: Faverage = 890 N a = 0.133 m/s2 10) this is the mass of both spacecraft so . . . m = 6692 kg