The Atwood machine has a 2 kg block on one side and 8 kg block on the other side.
Let us sketch a free-body diagram to better understand the problem.
Where F_tens is the tension in the string and F_grav is the force due to gravity.
The force due to gravity on the 2 kg block is given by
[tex]F_{2\operatorname{kg}}=m\cdot g=2\cdot9.81=19.62\; N[/tex]The force due to gravity on the 8 kg block is given by
[tex]F_{8\operatorname{kg}}=m\cdot g=8\cdot9.81=78.48\; N[/tex]According to Newton's second law of motion, we have
[tex]F_{\text{net}}=m\cdot a[/tex]The net force acting on the 2 kg block is given by
[tex]\begin{gathered} F_{tens}-F_{2\operatorname{kg}}=m\cdot a \\ F_{tens}-19.62=2\cdot a \\ F_{tens}=2a+19.62\quad eq.1 \end{gathered}[/tex]The net force acting on the 8 kg block is given by
[tex]\begin{gathered} F_{8\operatorname{kg}}-F_{\text{tens}}=m\cdot a \\ 78.48-F_{\text{tens}}=8a\quad eq.2 \end{gathered}[/tex]Now, we have two equations and two unknowns. Let us solve these two equations simultaneously
Substitute eq.1 into eq.2
[tex]\begin{gathered} 78.48-F_{\text{tens}}=8a \\ 78.48-(2a+19.62)_{}=8a \\ 78.48-2a-19.62_{}=8a \\ 78.48-19.62_{}=8a+2a \\ 58.86=10a \\ a=\frac{58.86}{10} \\ a=5.89\; \; \frac{m}{s^2} \end{gathered}[/tex]So, the acceleration of the two blocks is 5.89 m/s^2
Substitute a = 5.89 into eq. 1 to find the tension in the string.
[tex]\begin{gathered} F_{tens}=2a+19.62 \\ F_{tens}=2(5.89)+19.62 \\ F_{tens}=11.78+19.62 \\ F_{tens}=31.4\; N \end{gathered}[/tex]So, the tension of the string is 31.4 N
