Newton's First Law — Practice Problem Solutions

1. No. At the exact peak, the ball's velocity is momentarily zero, but its acceleration is still $-g$ — gravity is still acting, completely unbalanced. Equilibrium requires zero net force, not zero velocity, so the ball is not in equilibrium even at that instant.

2. $\sum F_y=0\implies T-mg=0\implies T=12\text{ N}$.

3. $\sum F_y=0\implies T+F_N-mg=0\implies T=mg-F_N=10-4=6\text{ N}$.

4. Yes, the puck is in equilibrium — constant velocity means zero acceleration, so the net force is zero. Forces are still acting on it (gravity down, normal force up) — they just balance exactly, and with no friction, nothing is needed to keep it moving at constant velocity (Newton's first law).

5. Horizontal balance: $F_1\cos\theta_1=F_2\cos\theta_2\implies F_1=\dfrac{F_2\cos\theta_2}{\cos\theta_1}$.

6. a) $F_{right}\cos\theta=F_{left}$. b) As $\theta$ decreases toward zero, $\cos\theta$ increases toward 1, so $F_{right}\cos\theta$ grows larger than the fixed $F_{left}$ — the horizontal balance breaks. The net horizontal force becomes positive in the direction of $F_{right}$, so the box speeds up (accelerates) in that direction instead of continuing at constant velocity.

7. a) $mg=Av_T^2$. b) $v_T=\sqrt{\dfrac{mg}{A}}$.

8. (B) — the elevator could possibly be moving downward and speeding up, or moving upward and slowing down. Both of those produce a net downward acceleration, which is exactly what makes the scale reading ($F_N$) come in below the true weight $mg$.