Question

Alguém me ajuda, não tô conseguindo ;c

Answer 1

$\underline{\text{Bloco M}}: \\\\ \text{F}_\text r=\text{P}_\text M-\text T \\\\ \text{M.a} = \text{M.g}-\text T\\\\\ \underline{\text{Bloco m}}: \\\\ \text{F}_\text r = \text T - \text{P}_\text m \\\\ \text{m.a}=\text T-\text{m.g} \\\\\\ \underline{\text{sistema}}: \\\\ \displaystyle \text{M.a} = \text{M.g}-\text T \\ \underline{\text{m.a}=\text T-\text{m.g}} \ \ + \\\\ \text a.(\text{M+m})= (\text M-\text m).\text g \\\\ \huge\boxed{\text{a}=\frac{(\text M-\text m).\text g}{(\text M+\text m)}\ }\checkmark$

Substituindo o valor da aceleração em uma das equações :

$\displaystyle \text {m.a} = \text T - \text{m.g} \\\\ \text T = \text{m.g}+\text{m.a} \\\\ \text T = \text{m.g}+\frac{\text m.(\text M-\text m).\text g}{(\text M+\text m)} \\\\\\ \text T = \frac{\text{m.g}.(\text M+\text m)+\text m.(\text M-\text m).\text g}{(\text M+\text m)} \\\\\\\ \text T = \frac{\text{m.M.g}+\text m^2.\text g+\text{m.M.g}-\text m^2.\text g}{(\text M+\text m)} \\\\\\ \huge\boxed{\ \text T = \frac{\text{2.m.M.g}}{(\text M+\text m)}\ }\checkmark$