Two bodies of masses m_1 = 2 kg and m_2 = 3 kg , on a…

Two bodies of masses $m_1 = 2\ \mathrm{kg}$ and $m_2 = 3\ \mathrm{kg}$, on a frictionless horizontal surface, are connected by a rope that breaks at a tension of $T_{\max} = 350\ \mathrm{N}$. Forces growing with time, $F_1 = kt$ on $m_1$ and $F_2 = 2kt$ on $m_2$, with $k = 10\ \mathrm{N/s}$, pull the bodies in opposite directions (apart). When does the rope break?

Explanation of Method

When a rope between two pulled bodies breaks.

Step 1: Acceleration of the pair

$F_2$ wins, so both move in its direction: $a = \frac{F_2 - F_1}{m_1 + m_2} = \frac{kt}{5}$.

Step 2: Tension

For $m_1$, the rope pulls forward and $F_1$ backward: \[T - F_1 = m_1 a \;\Longrightarrow\; T = kt + \frac{2kt}{5} = \frac{7}{5}kt = 14t\ \mathrm{N}.\]

Step 3: Breaking time

\[ 14\,t = 350. \]

\[t = 25\ \mathrm{s}\]

Check

For $m_2$: $ F_2 - T = 2kt - \tfrac75 kt = \tfrac35 kt = m_2 \cdot \tfrac{kt}{5}\ \checkmark $

$F_1 = kt$ $F_2 = 2kt$ $2$ $3$ $T$ $T\ (\mathrm{N})$ $t\ (\mathrm{s})$ $0$ $10$ $20$ $25$ $30$ $350$ $\text{breaks at } 25\ \mathrm{s}$ $T = \tfrac{7}{5}kt$

Forces pulling the bodies apart (assumed arrangement). The rope must hold back $m_1$ against $F_1$ and also drag it along with the system, so $T = F_1 + m_1a$ grows like $\tfrac75 kt$.

Result

\[\boxed{\,t = 25\ \mathrm{s}\,}\]

Remark

About the original solution. It added the two forces as if they pulled the same way and then wrote $T = m_1 a$, forgetting that $F_1$ also acts on $m_1$. That gives 29.17 s, which is not right for either arrangement. With the forces pulling the bodies apart (the usual setup for breaking a rope), $t = 25\ \mathrm{s}$. If both forces pulled in the same direction ($F_2$ on the front body), the tension would be only $T = \tfrac15 kt = 2t$ and the rope would break at $t = 175\ \mathrm{s}$.