Newton's Second Law (Part 1) — Practice Problem Solutions

Using $g=9.8\text{ m/s}^2$ unless otherwise noted.

1. $F_N-mg=ma\implies F_N=m(g+a)=50(9.8+1)=540\text{ N}$.

2. $a=g\sin\theta$ (down the ramp).

3. $F_N=m(g-a)=70(9.8-3)=476\text{ N}$.

4. $F_{net}=ma=m(Ct)=mCt$ — a straight line through the origin with slope $mC$, increasing linearly with time.

5. The horizontal push's component along the ramp is $F_p\cos\theta$ (up the ramp); gravity's component along the ramp is $mg\sin\theta$ (down the ramp). $a=\dfrac{F_p\cos\theta-mg\sin\theta}{m}$.

6. a) At $t_1$: $F_N=m(g-a)=60(9.8-3)=408\text{ N}$. At $t_2$: $F_N=m(g+a)=60(9.8+1)=648\text{ N}$. b) $\dfrac{408}{648}\approx0.63$.

7. a) The spacecraft's acceleration increases over time. Since $a=\dfrac{F_{net}}{m}$ and the thrust force stays constant while mass steadily decreases, the same force acting on less and less mass produces more and more acceleration. b) The relationship isn't linear because acceleration depends on $\dfrac{1}{m}$, not on $m$ directly — even if the mass itself decreases at a steady rate, $\dfrac{1}{m(t)}$ is not a straight-line function of time, so the acceleration curves upward rather than climbing steadily.

8. (Using $g=10\text{ m/s}^2$ as given.) $F_N-mg=ma\implies a=\dfrac{1100-900}{90}\approx2.2\text{ m/s}^2$, directed upward. (C) — the elevator could be moving upward and speeding up, or moving downward and slowing down; both situations produce the same net upward acceleration that this scale reading requires.