Newton's Second Law (Part 2: Connected Systems) — Practice Problem Solutions

1. Treat the two blocks as one system: $a=\dfrac{F}{m_{total}}=\dfrac{16}{3+5}=2\text{ m/s}^2$.

2. Isolating the trailing 5 kg block, tension is the only horizontal force on it: $T=ma=(5)(2)=10\text{ N}$.

3. No — the center of mass is not accelerating. With no net external force acting on the two-object system (their mutual gravity is an internal force pair that cancels when you add up the whole system), the center of mass moves at constant velocity, even though each individual object is undergoing accelerating (circular) motion of its own.

4. Whole-system equation: $m_2g=(m_1+m_2)a\implies a=\dfrac{m_2g}{m_1+m_2}$.

5. Isolating $m_1$: $T=m_1a=\dfrac{m_1m_2g}{m_1+m_2}$.

6. a) With the string broken, Crate 2 only has its own friction acting on it: $a_2=-\dfrac{f_2}{m}$ (a deceleration). b) The system's center-of-mass acceleration $=\dfrac{F-f_1-f_2}{3m}$ (total external force divided by total mass). This can differ from Crate 2's own acceleration because Crate 1 is still being driven forward by $F$ and continues to speed up even as Crate 2 slows down — the center-of-mass acceleration blends both objects' individual motions (weighted by mass), so it doesn't have to match what's happening to either crate alone.

7. a) No, the net force is not zero — since $a_{cm}=\dfrac{F_{net}}{m_{total}}$, a nonzero acceleration means the net force must be nonzero too. b) Equilibrium requires zero net force (and therefore zero acceleration), not just zero velocity. Here the system has a momentary zero velocity but a nonzero acceleration — it's about to start moving, the opposite of sitting in equilibrium. (Compare to a ball thrown straight up: it has zero velocity at its peak, but it's still accelerating due to gravity.)

8. Isolating Crate 2 (the trailing block, with only tension acting on it horizontally): $F_T=m_2a$ — (C).