Physics - NLM Question with Solution | TestHub

Figure (a): $F-T_1=m_{1}a\Rightarrow 1000-T_1=50a$; $T_1-m_{2}g=m_{2}a\Rightarrow T_1-800=80a$ (taking $g=10$). So $T_1=\frac{12000}{13}\mathrm{N}$.

Figure (b): the force on $m_1$ is $F+m_{1}g$ (with $m_{1}g=500\mathrm{N}$ in the direction of motion): $1000+500-T_2=50a$; $T_2-800=80a$. So $T_2=\frac{16000}{13}\mathrm{N}$.

$T_1:T_2=12000:16000=3:4$.