If mass A is 1.0 kg, mass B is 5.0 kg, and the frictional pulley has a mass of 0.5 kg and a radius of 0.15 m, what is the velocity of A and B at 0.5 seconds? a. A = 1.72 m/s, B = 0.22 m/s
c. A = 0.22 m/s, B = 1.72 m/s
b. A = 3.14 m/s, B = -3.14 m/s
d. A = -3.14 m/s, B = 3.14 m/s

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.

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explanation:

an energy transformation flow diagram is shown

where is it ?