| Derivation / Formula | Reasoning |
|---|---|
| \[F_{\text{net, applied}} = 5\, \text{N (right)} – 3\, \text{N (left)} = 2\, \text{N (right)}\] | The two external pushes combine to give a net applied force of \(2\,\text{N}\) toward the right. |
| \[F_{\text{net, applied}} + f = 0\] | Because the block moves at constant velocity, Newton’s First Law requires the total horizontal force to be zero; friction \(f\) must balance the applied forces. |
| \[f = -2\,\text{N}\] | The negative sign shows friction acts opposite the net applied force—that is, toward the left. |
| \[|f| = 2\,\text{N},\; \text{leftward}\] | Magnitude and direction of friction: \(2\,\text{N}\) to the left. |
| Why others are wrong: (a) With no friction, the unbalanced \(2\,\text{N}\) would accelerate the block. (c) A rightward \(2\,\text{N}\) friction force would double the net rightward force. (d) & (e) A \(7\,\text{N}\) friction force over- or under-compensates, again giving a non-zero net force, violating the constant-velocity condition. |
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A cart with a mass of \( 20 \) \( \text{kg} \) is pressed against a wall by a horizontal spring with spring constant \( k = 244 \) \( \text{N/m} \) placed between the cart and the wall. The spring is compressed by \( 0.1 \) \( \text{m} \). While the spring is compressed, an additional constant horizontal force of \( 20 \) \( \text{N} \) continues to push the cart toward the wall. What is the resulting acceleration of the cart?
In the diagram above, block \(A\) has a mass of \(3.2 \, \text{kg}\) and block \(B\) a mass of \(2.4 \, \text{kg}\). The pulley is frictionless and has no mass.
A space probe far from the Earth is travelling at \( 14.8 \) \( \text{km s}^{-1} \). It has mass \( 1\,312 \) \( \text{kg} \). The probe fires its rockets to give a constant thrust of \( 156 \) \( \text{kN} \) for \( 220. \) \( \text{s} \). It accelerates in the same direction as its initial velocity. In this time it burns \( 150. \) \( \text{kg} \) of fuel.
Calculate the final speed of the space probe in \( \text{km s}^{-1} \).
The Earth’s radius is \(6.37 \times 10^{6} \, \text{m}\). What is the radius of a planet that has the same mass as Earth but on which the free-fall acceleration is \(5.50 \, \text{m/s}^2\)?
A constant force of 8.0 N is exerted on a 16 kg object initially at rest. How much speed will the object gain after 4 seconds?
Why is more fuel required for a spacecraft to travel from the Earth to the Moon than to return from the Moon to the Earth?
The gravitational force that the Moon exerts on Earth is often cited as the source of the tides we witness. However, the gravitational force the Sun exerts on Earth is over \(100\) times greater than the force the Moon exerts on Earth.
Why is the force from the Moon credited for the tides, and not the force from the Sun?
A box is sliding down an incline at a constant speed of \( 2 \) \( \text{m s}^{-1} \). The angle of the incline is \( \theta \). The magnitude of the total of the opposing forces is \( 16 \) \( \text{N} \). What is the force of gravity acting on the box?
If an elephant were chasing you, its enormous mass would be most threatening. But if you zigzagged, its mass would be to your advantage. Why?
A block starts from rest at the top of a \(50^\circ\) incline. The coefficient of kinetic friction between the block and the incline is \(0.4\). If the block reaches a velocity of \(7 \, \text{m/s}\) at the bottom of the incline, what is the length of the incline?
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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] |
Metric Prefixes
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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