Statics and Dynamics Quiz: Newtons Second Law Cartesian

What this quiz covers

This quiz focuses on Newtons Second Law Cartesian, giving you a quick way to practice the rules, question types, and explanations that matter most for Statics and Dynamics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $x$: $m\dot{v}_x = \mu_k mg$; $y$: $N - mg = 0$. Slipping lasts $t = v_b/(\mu_k g) \approx 1.02\text{ s}$, after which the package reaches belt speed; the friction force is positive in $+x$ during slip because the belt moves faster than the package. (correct answer)
2. $x$: $m\dot{v}_x = -\mu_k mg$; $y$: $N - mg = 0$. Slipping lasts $t = v_b/\mu_k \approx 10\text{ s}$, because the friction force opposes belt motion and decelerates the package until it matches belt speed.
3. $x$: $m\dot{v}_x = \mu_k N$; $y$: $N - mg = m\dot{v}_y$. Slipping lasts $t = v_b/(\mu_k g) \approx 1.02\text{ s}$, but the $y$-equation shows $N > mg$ during slipping because the friction force has a vertical reaction component.
4. $x$: $m\dot{v}_x = \mu_k mg$; $y$: $N - mg = 0$. Slipping lasts $t = v_b/(2\mu_k g) \approx 0.51\text{ s}$, because the average relative velocity during slip is half the belt speed, reducing the effective slipping duration by a factor of two.

Question 2

1. Use $m\ddot{x} = F_w$ and $m\ddot{y} = -mg$ for the entire flight, integrate both equations from $t = 0$ to the landing time $t_f$ found from $y(t_f) = 0$; the wind force can be treated as constant throughout because the particle necessarily passes back through $y < H$ on its descent.
2. Split into two phases at the apex $y = y_{max}$: (1) from launch to apex with $m\ddot{x} = F_w$, $m\ddot{y} = -mg$; (2) from apex to landing with $m\ddot{x} = 0$, $m\ddot{y} = -mg$; the wind ceases at the apex because the particle stops moving in $y$, which triggers the altitude condition.
3. Split the trajectory into three phases: (1) ascent with $y < H$: $m\ddot{x} = F_w$, $m\ddot{y} = -mg$; (2) above $H$: $m\ddot{x} = 0$, $m\ddot{y} = -mg$; (3) descent with $y < H$: $m\ddot{x} = F_w$, $m\ddot{y} = -mg$; match velocity and position at each phase boundary to find total range. (correct answer)
4. Use $m\ddot{x} = F_w$ and $m\ddot{y} = -mg$ while $y < H$, then $m\ddot{x} = 0$ and $m\ddot{y} = -mg$ while $y \geq H$; split into two phases at the moment the particle crosses $y = H$ on its way up, and assume it does not re-enter $y < H$ until landing, ignoring the descending portion below $H$.

Question 3

1. The $x$-motion is oscillatory (simple harmonic about $x = 0$) and bounded, while the $y$-motion diverges exponentially; the particle is unbounded because the $x$-force is restoring but the $y$-force repels from $y = 0$, and initial conditions in $y$ cause exponential growth.
2. Both the $x$- and $y$-motions are simple harmonic and bounded because both force components are linear in displacement; the particle executes Lissajous motion and remains bounded for all time.
3. The $x$-motion diverges exponentially away from $x = 0$ and is unbounded, while the $y$-motion is simple harmonic and bounded; the particle is unbounded because the $x$-force $F_x = 12x$ is a repulsive (unstable) force proportional to displacement. (correct answer)
4. The $x$-motion is simple harmonic and bounded, while the $y$-motion diverges; the particle is bounded overall because the $y$-direction energy is transferred to the $x$-direction through the coupled equations of motion, preventing unbounded growth.

Question 4

1. $\vec{F}_3 = F_0(1 - \cos\theta)\hat{i} - F_0\sin\theta\hat{j} + ma_0\hat{j}$; the constraint is $\sin\theta = ma_0/F_0$, ensuring the $\hat{j}$ component of $\vec{F}_3$ vanishes while the block still accelerates in $\hat{j}$.
2. $\vec{F}_3 = F_0(1 - \cos\theta)\hat{i} + (ma_0 - F_0\sin\theta)\hat{j}$; the constraint $\sin\theta = ma_0/F_0$ eliminates the $\hat{j}$ component of $\vec{F}_3$, so the required $\hat{j}$ acceleration comes entirely from $\vec{F}_1$'s vertical component. (correct answer)
3. $\vec{F}_3 = F_0\cos\theta\hat{i} + (ma_0 - F_0\sin\theta)\hat{j}$; the constraint $\sin\theta = ma_0/F_0$ eliminates the $\hat{j}$ component of $\vec{F}_3$, with the $\hat{i}$ component of $\vec{F}_3$ balancing the net horizontal forces from $\vec{F}_1$ and $\vec{F}_2$ together.
4. $\vec{F}_3 = -F_0(1 - \cos\theta)\hat{i} + (ma_0 - F_0\sin\theta)\hat{j}$; the constraint $\theta = \pi/2$ is the only angle ensuring $\vec{F}_3$ has no $\hat{j}$ component when $F_0 = ma_0$, since all horizontal forces cancel at that angle.

Question 5

1. Differentiating $m\ddot{x} = ay$ twice gives $m\ddddot{x} = a\ddot{y}$; substituting $\ddot{y} = \frac{b}{m}x$ yields $\ddddot{x} = \frac{ab}{m^2}x$, but since $a = b$ the equation reduces to $\ddddot{x} = \frac{a^2}{m^2}x$ with only real exponential solutions $e^{\pm(a/m)t}$, because the equal coefficients eliminate the imaginary roots.
2. Differentiating $m\ddot{x} = ay$ once gives $m\dddot{x} = a\dot{y}$; this cannot be further reduced because substituting the $y$-equation introduces $\dot{x}$ terms, making the system irreducibly coupled and requiring numerical integration.
3. Differentiating $m\ddot{x} = ay$ twice and substituting from the $y$-equation yields $m^2\ddddot{x} = ab\,\ddot{x}$; setting $\omega^2 = ab/m^2$ gives a characteristic equation $r^4 = \omega^2 r^2$, producing only oscillatory and zero roots with general solution $x(t) = C_1 + C_2 t + C_3\cos(\omega t) + C_4\sin(\omega t)$.
4. Differentiating $m\ddot{x} = ay$ twice and substituting $\ddot{y} = \frac{b}{m}x$ gives $\ddddot{x} = \frac{ab}{m^2}x$; with $a = b = 6$, $m = 2$ this becomes $\ddddot{x} = 9x$, whose characteristic roots are $r = \pm\sqrt{3},\,\pm i\sqrt{3}$, yielding $x(t) = C_1 e^{\sqrt{3}t} + C_2 e^{-\sqrt{3}t} + C_3\cos(\sqrt{3}t) + C_4\sin(\sqrt{3}t)$. (correct answer)

Question 6

1. The trajectory curves toward the $x$-axis as $t$ increases, eventually becoming parallel to $\hat{i}$, because the $y$-component of velocity decays faster than the $x$-component due to the quadratic nature of the drag coupling between components.
2. The trajectory is a straight line in the direction of $\vec{v}_0$ for all $t$, and the speed decays exponentially; the direction of motion remains $\alpha$ from the $x$-axis throughout the motion because the drag force components are proportional to velocity components independently. (correct answer)
3. The trajectory spirals inward toward the origin because the drag force has both a speed-reducing component and a direction-changing component that causes the heading angle $\alpha$ to decrease monotonically over time.
4. The trajectory is a straight line only if $\alpha = 0$ or $\alpha = \pi/2$; for other launch angles the drag coupling between components causes the path to curve toward the direction of the larger initial velocity component.