La tierra aplica una fuerza de atracción de 1000 N sobre un satélite de comunicaciones. La masa del satélite es más de un millón
1 answer:
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Answer
given,
mass of the person, m = 50 Kg
length of scaffold = 6 m
mass of scaffold, M= 70 Kg
distance of person standing from one end = 1.5 m
Tension in the vertical rope = ?
now equating all the vertical forces acting in the system.
T₁ + T₂ = m g + M g
T₁ + T₂ = 50 x 9.8 + 70 x 9.8
T₁ + T₂ = 1176...........(1)
system is equilibrium so, the moment along the system will also be zero.
taking moment about rope with tension T₂.
now,
T₁ x 6 - mg x (6-1.5) - M g x 3 = 0
'3 m' is used because the weight of the scaffold pass through center of gravity.
6 T₁ = 50 x 9.8 x 4.5 + 70 x 9.8 x 3
6 T₁ = 4263
T₁ = 710.5 N
from equation (1)
T₂ = 1176 - 710.5
T₂ = 465.5 N
hence, T₁ = 710.5 N and T₂ = 465.5 N
Answer:
v = 8.45 m/s
Explanation:
given,
mass = 3 kg
angle = 30.0°
vertical distance = 3.3 m
μ = 0.06
according to conservation of energy
KE(loss) = PE(gain) + Work done (against\ friction)..............(1)
frictional Force
work against friction
W = F d
Potential energy
PE = mgh
v = 8.45 m/s
the minimum speed is equal to 8.45 m/s