Atps fisica ii primeira parte

Resolução
13)

P = m*g
P = 8.5*9.8
P = 83.3 N

N = Py N = 72.13 (N) (Py = Cos 30º*83.3)
Px = P*Senθ
Px = 83.3*Sen30º
Px = 41.65 N

T = Px
T = 41.65 N

Py = P.Senθ
Py = 83.3*Cos30º
Py = 72.13 N

a = Px/m a = 41.65/8.5 a = 4,9 m/s²

Pg. 16 – a)

a=10m/s² 20 = 20-F2 a=∆F/m F2=0N a=(10-F2)/2 b)

a=20m/s² 2a=(20-F2)/2 a=∆F/m 40=20-F2 a=(F1-F2)/m F2=20N

c) a=0m/s² a=∆F/m a= (F1-F2) a = 20-F2 a =20 N

d)

a= - 10m/s² a=∆F/m

a=(F1-F2)/m -10 = (20- F1)/2 F2 = 40 N

e) a= 20m/s² a=∆F/m

a=(F1-F2)/m -20 = (20- F2)/2 -40 = 20 – F2 F2 = 60 N

Pg. 119 – 49 – a)

a=Fx/m a = (F Cos25º)/5 a = (12 Cos25º)/5 a = 2,18 m/s²

b) F Sem 25º + N – m*g = 0
F = (m*g)/(Sen 25) => F = (5*9.8)/(Sen 25) => F = 116 N

e) a = (F Cos 25º)/m => a = (116 Cos 25º)/5 => a = 21,02 m/s²

120 – 49 – a)

a = (F Mm*g)/Mm => a = ((15-10)*9.8)/10 => a = 4,9 m/s²

b) am = ((Mρ-Mm)*g)/(Mρ-Mm) => a = 2 m/s²

c) F=mp (g-am) => F=15*(9.8-2) => F=117 N

145 – 29 Bloco m Bloco M Bloco m
∑ Fx = m*a*x
F – Nm = m*a
F = m*a + Nm

∑Fy = 0
Pm – fe = m*g – We*Nm = 0
Nm = (m*g)/We

Bloco M

Nm = Ma a = N/M
Substituindo 2 e 3 em 1

F= m (m*g)/(M*We) + (m*g)/(M*We) = (m*g)/We ( (mg )/Me+1)
F= (16*9.8)/0,38 ( (16 )/88+1) = 486,9 ≈ 4,86*10²

Pg- 119 – 39

MA
Fr=0
Fr = m*a
Fr = 0.10*2.5
Fr = 0,25

MB
Fr = 0
Fr = m*a
Fr = 0,25*2,5
Fr = 0,62 N

MC
Fr = 0
Fr = m*a
Fr = 0,62*2,5
Fr = 1,56 N

MD
Fr = 0
Fr =