In the system shown below, friction and mass of the pulley are negligible. Find the acceleration (in $\mathrm{m}/{\mathrm{s}}^2$ ) of m₂ if ${\mathrm{m}}_1=300\mathrm{g},{\mathrm{m}}_2=500\mathrm{g}$ and F = 3.4 N

When the pulley moves a distance d, m₁ will move a distance 2d. Hence m₂ will have twice as large an acceleration as m₂ has. 

Also because the total force on the pulley must be zero,T₁ = (T₂/2)

For mass m₁, T₁ = m₁ (2a)  … (i)

 For mass m₂, F – T₂ = m₂(A) … (ii)
Putting T₁ = $\frac{{\mathrm{T}}_2}{2}$  , (i) gives T₂ = 4m₁a 

Substituting in equation (ii), F = 4m₁a + m₂a = (4m₁ + m₂)a 

Hence $\mathrm{a}=\frac{\mathrm{F}}{\left(4{\mathrm{m}}_1+{\mathrm{m}}_2\right)}=\frac{3.4}{4(0.3)+0.5}=2\mathrm{m}/{\mathrm{s}}^2$