Three blocks with masses m, 2m and 3m are connected by strings, as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed $\upsilon$. What is the net force on the block of mass 2m? (g is the acceleration due to gravity)

Let $T_1$ be tension in string connecting m and 2m and $T_2$ be tension in string connecting 2m and 3m.

            Let a be common acceleration of the system.

$\therefore a=\frac{F-\left(m+2m+3m\right)g}{m+2m+3m}=\frac{F-6m}{6m}$

As the system moves with constant speed, therefore, a = 0

$\therefore F-6mg=0\Rightarrow F=6mg$

The free body diagram of block m is as shown in the figure.

The equation of motion of block of mass m is

$$\begin{gathered} F-T_1-mg=0 \\ \Rightarrow 6mg-T_1-mg=0 \\ \Rightarrow T_1=5mg---\left(i\right) ; \end{gathered}$$

The free body diagram of block of mass 2m is as shown in the figure.

The equation of motion of block of mass 2m is

$$\begin{gathered} {\mathrm{T}}_1-{\mathrm{T}}_2-2\mathrm{mg}=0 \\ \Rightarrow 5\mathrm{mg}-{\mathrm{T}}_2-2\mathrm{mg}=0\hspace{0.17em} \left(\mathrm{using}\hspace{0.17em}\hspace{0.17em}\left(\mathrm{i}\right)\right) \\ \Rightarrow {\mathrm{T}}_2=3\mathrm{mg} \end{gathered}$$

The free body diagram of block of mass 3m is as shown in the figure.

The equation of motion of block of mass 3m is

$$\begin{gathered} T_2-3mg=0 \\ \Rightarrow T_2=3mg \end{gathered}$$

Net force on the block of mass 2m is

$$\begin{gathered} F_{net}=T_1-T_2-2mg \\ =5mg-3mg-2mg=0 \end{gathered}$$

Alternate solution

As all blocks are moving with constant speed, therefore, acceleration is zero. So net force on each block is zero.