First, let's calculate the forces in the 3 kg mass:
[tex]W_1-T=m_1\cdot a[/tex]
Then, the forces in the 2 kg mass:
[tex]T-W_2=m_2\cdot a[/tex]
From the first equation, let's solve it for T:
[tex]T=W_1-m_1a[/tex]
Using this value of T in the second equation, we have:
[tex]\begin{gathered} W_1-m_1a-W_2=m_2\cdot a \\ W_1-W_2=m_2a+m_1a \\ m_1g-m_2g=a\cdot(m_1+m_2)_{} \\ a=g\frac{(m_1-m_2)}{(m_1+m_2)} \end{gathered}[/tex]
Finally, using the given values (with g = 10 m/s²), we have:
[tex]a=10\cdot\frac{(3-2)}{(3+2)}=10\cdot\frac{1}{5}=2\text{ m/s2}[/tex]
If the acceleration is 2 m/s², after 2 seconds the speed is:
[tex]\begin{gathered} V=V_0+a\cdot t \\ V=0+2\cdot2 \\ V=4\text{ m/s} \end{gathered}[/tex]
Therefore the correct option is the second one.
