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Let’s analyze the forces acting on the ball and understand how they affect the acceleration of the ball at time \(t/2\).
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( F_{\text{net}} = ma = mg – bv \) | The net force (\(F_{\text{net}}\)) acting on the ball is the difference between the gravitational force (\(mg\)) and the retarding force (\(F = bv\)). Using Newton’s second law, this net force is also equal to \(ma\). |
| 2 | \( a = g – \frac{bv}{m} \) | Solving for acceleration (\(a\)), we rearrange the equation \(F_{\text{net}} = ma\) and isolate \(a\). |
| 3 | At \(t = \frac{t}{2}\), \(v < v_{\text{terminal}}\) | At time \(t/2\), the ball hasn’t yet reached terminal velocity, so the speed \(v\) is less than the terminal velocity (\(v_{\text{terminal}}\)). |
| 4 | \( v_{\text{terminal}} = \frac{mg}{b} \) | The terminal velocity is reached when the retarding force equals the gravitational force, i.e., \(bv = mg\). Solving for \(v\) gives the terminal velocity. |
| 5 | \( a_{\frac{t}{2}} = g – \frac{bv}{m} \), \( v < v_{\text{terminal}} \Rightarrow \frac{bv}{m} < g \) | At \(t/2\), since \(v < v_{\text{terminal}}\), the fraction \(\frac{bv}{m}\) is less than \(g\), which implies a positive acceleration. The magnitude of the acceleration is decreasing as it has not reached zero yet. |
| 6 | Magnitude of acceleration is decreasing | The acceleration decreases because the retarding force increases with speed, reducing the net force and thereby the acceleration over time. |
Final Conclusion: At time \(t/2\), the magnitude of the acceleration of the ball is decreasing. Therefore, the correct choice would be: (b) Decreasing. The incorrect choices: (a) Increasing would be incorrect as acceleration decreases over time due to the velocity-dependent resistive force; (c) 10 m/s² is incorrect because the acceleration would be less than \(g=10\text{ m/s}^2\) due to the retarding force; (d) Zero is incorrect because the ball has not yet reached terminal velocity at \(t/2\).
Just ask: "Help me solve this problem."
A block sliding down an frictionless inclined plane is experiencing both gravitational and normal forces; which force’s magnitude changes when the angle of the incline is increased?
An \( 80 \, \text{kg} \) block is placed \( 2 \, \text{m} \) away from the endpoint of a horizontal steel beam of length \( 6.6 \, \text{m} \) and mass \( 1,450 \, \text{kg} \). The plank makes contact with a vertical wall on one end, and the other endpoint is attached to a massless cable that makes an angle of \( 30^\circ \) with the horizontal and ties into the vertical wall as well. Calculate the magnitude and direction of the force between the cable and the wall and of the force between the steel beam and the wall.
In the diagram shown a 20 N force is applied to a block B (7 kg). Block A has a mass of 3 kg. Assume frictionless conditions.
A horizontal, uniform board of weight \( 125 \, \text{N} \) and length \( 4 \, \text{m} \) is supported by vertical chains at each end. A person weighing \( 500 \, \text{N} \) is hanging from the board. The tension in the right chain is \( 250 \, \text{N} \).
A 10kg box is pushed to the right by an unknown force at an angle of 25° below the horizontal while a friction force of 50 N acts on the box as well. The box accelerates from rest and travels a distance of 4 m where it is moving at 3 m/s. Solve the following without the use of energy.
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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