Three masses are arranged on a table with pulleys, as shown. the pulleys and string are taken to be massless. there is a coefficient of kinetic friction μk = 0.2 between m3 and the table. if m1 =4kg, m3=6kg and m2 =12kg, then the acceleration of the masses is
the figure looks like a table with a box in the middle attached to string and two pulleys on either side with two other boxes hanging off the table.

3.03 m/s²

Explanation:

m₂ > m₁, so m₁ will move up, m₂ will move down, and m₃ will move to the right.

Sum of the forces on m₁:

∑F = ma

T₁ − m₁g = m₁a

T₁ = m₁g + m₁a

Sum of the forces on m₂:

∑F = ma

T₂ − m₂g = m₂(-a)

T₂ = m₂g − m₂a

Sum of the forces on m₃ in the y direction:

∑F = ma

N − m₃g = 0

N = m₃g

Sum of the forces on m₃ in the x direction:

∑F = ma

T₂ − T₁ − F = m₃a

T₂ − T₁ − Nμ = m₃a

Substituting and solving for acceleration:

(m₂g − m₂a) − (m₁g + m₁a) − (m₃g)μ = m₃a

m₂g − m₂a − m₁g − m₁a − m₃gμ = m₃a

g (m₂ − m₁ − m₃μ) = (m₁ + m₂ + m₃) a

a = g (m₂ − m₁ − m₃μ) / (m₁ + m₂ + m₃)

Given m₁ = 4 kg, m₂ = 12 kg, m₃ = 6 kg, and μ = 0.2:

a = 9.81 (12 − 4 − 6×0.2) / (4 + 12 + 6)

a = 3.03 m/s²

depends on what you mean by "group." for order, it would be coleoptera (beetles). for class, it would be insecta (insects). for phylum, it would be the arthropoda (animals with jointed exoskeletons).

explanation: