Two bodies of masses m_1 = 10 kg and m_2 = 20 kg , connected…

Two bodies of masses $m_1 = 10\ \mathrm{kg}$ and $m_2 = 20\ \mathrm{kg}$, connected by a rope, lie on a horizontal surface. The body $m_2$ is pulled with a force $F = 20\ \mathrm{N}$. Find the acceleration and the rope tension. Neglect friction.

Explanation of Method

Two bodies joined by a rope.

Step 1: Acceleration

\[ a = \frac{F}{m_1 + m_2} = \frac{20}{30}. \]

\[a = \tfrac23\ \mathrm{m/s^2} \approx 0.667\ \mathrm{m/s^2}\]

Step 2: Tension

The rope is the only horizontal force on $m_1$: $T = m_1 a = \tfrac{20}{3}\ \mathrm{N} \approx 6.67\ \mathrm{N}$.

Check

For $m_2$: $ F - T = 20 - 6.67 = 13.33\ \mathrm{N} = 20 \cdot \tfrac23\ \checkmark $

$F = 20\ \mathrm{N}$ $T = 6.67\ \mathrm{N}$ $10\ \mathrm{kg}$ $20\ \mathrm{kg}$ $a = \frac{F}{m_1 + m_2} = \frac{2}{3}\ \mathrm{m/s^2}$

The rope only has to accelerate the rear body, so it carries just a third of the pulling force.

Result

\[\boxed{\,a = \tfrac23\ \mathrm{m/s^2} \approx 0.667\ \mathrm{m/s^2}\,}\]

\[\boxed{\,T = m_1 a = \tfrac{20}{3}\ \mathrm{N} \approx 6.67\ \mathrm{N}\,}\]