Answer:
b. A = 3.14 m/s, B = -3.14 m/s
Explanation:
Assuming this refers to an Atwood machine, draw free body diagrams for each mass.
For mass A, there are two forces: tension T₁ pulling up and weight m₁g pulling down.
For mass B, there are two forces: tension T₂ pulling up and weight m₂g pulling down.
For the pulley, there are two torques: tension T₁r pulling counterclockwise and tension T₂r pulling clockwise.
Sum of forces on A in the +y direction:
∑F = ma
T₁ − m₁g = m₁a
T₁ = m₁g + m₁a
Sum of forces on B in the -y direction:
∑F = ma
m₂g − T₂ = m₂a
T₂ = m₂g − m₂a
Sum of torques on the pulley in the clockwise direction:
∑τ = Iα
T₂r − T₁r = (½ mr²) (a/r)
T₂ − T₁ = ½ ma
Substitute:
m₂g − m₂a − (m₁g + m₁a) = ½ ma
m₂g − m₂a − m₁g − m₁a = ½ ma
m₂g − m₁g = m₂a + m₁a + ½ ma
g (m₂ − m₁) = a (m₂ + m₁ + ½ m)
a = g (m₂ − m₁) / (m₂ + m₁ + ½ m)
a = (9.8 m/s²) (5.0 kg − 1.0 kg) / (5.0 kg + 1.0 kg + ½ (0.5 kg))
a = 6.272 m/s²
Given:
v₀ = 0 m/s
a = 6.272 m/s²
t = 0.5 s
Find: v
v = at + v₀
v = (6.272 m/s²) (0.5 s) + 0 m/s
v = 3.14 m/s
Therefore, mass A will rise at 3.14 m/s, and mass B will fall at 3.14 m/s.
