| Derivation or Formula | Reasoning |
|---|---|
| \[g_p = \frac{G M_p}{R_p^2}\] | Definition of the surface gravitational field on the new planet, where \(g_p = 20\,\text{N/kg}\). |
| \[g_E = \frac{G M_E}{R_E^2}\] | Surface gravitational field on Earth, with \(g_E \approx 9.8\,\text{N/kg}\). |
| \[\frac{g_p}{g_E} = \frac{G M_p / R_p^2}{G M_E / R_E^2} = \frac{M_p}{M_E}\,\frac{R_E^2}{R_p^2}\] | Taking the ratio of the two gravitational field equations cancels the universal gravitational constant \(G\). |
| \[\frac{20}{9.8} = \frac{0.5\,R_E^2}{R_p^2}\] | The planet’s mass is half of Earth’s, so \(M_p = 0.5\,M_E\). |
| \[R_p^2 = 0.5 \left(\frac{9.8}{20}\right) R_E^2 = 0.245\,R_E^2\] | Algebraically solve for \(R_p^2\). |
| \[R_p = \sqrt{0.245}\,R_E \approx 0.495\,R_E\] | Take the square root to find \(R_p\) in terms of \(R_E\). |
| \[R_p \approx 0.495 \times 6400\,\text{km} \approx 3.17 \times 10^3\,\text{km}\] | Substitute \(R_E = 6400\,\text{km}\) to get the numerical value. |
| \[\boxed{R_p \approx 3.2 \times 10^3\,\text{km}}\] | Final radius of the new planet. |
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Three blocks of masses \( 1.0 \, \text{kg} \), \( 2.0 \, \text{kg} \), and \( 4.0 \, \text{kg} \) are connected by massless strings, one of which passes over a frictionless pulley of negligible mass, as shown above. Calculate each of the following.
Two closed containers look the same, but one is packed with lead and the other with a few feathers. How could you determine which has more mass if you and the containers were orbiting in a weightless condition in outer space?
The occupants of a car traveling at a speed of \( 30 \) \( \text{m/s} \) note that on a particular part of a road their apparent weight is \( 15\% \) higher than their weight when driving on a flat road.
A \( 60 \ \text{kg} \) person is riding in an elevator. At time \( t_1 \), the elevator is accelerating downward with a magnitude of \( 2 \ \text{m/s}^2 \). A short time later, at time \( t_2 \), the elevator is accelerating upward with a magnitude of \( 2 \ \text{m/s}^2 \). The ratio of the normal force exerted by the elevator on the person at time \( t_1 \) to that at time \( t_2 \) is most nearly
When the speed of a rear-wheel-drive car is increasing on a horizontal road, what is the direction of the frictional force on the tires?
A point \( P \) is subjected to three simultaneous forces of magnitudes \( F_A > F_B > F_C \). Point \( P \) is in equilibrium. Which of the following statements is not always true about the magnitudes of the forces?
A spring with a spring constant of \( 50. \) \( \text{N/m} \) is hanging from a stand. A second spring with a spring constant of \( 100. \) \( \text{N/m} \) is hanging from the first spring. How far do they stretch if a \( 0.50 \) \( \text{kg} \) mass is hung from the bottom spring?
A \(25.0 \, \text{kg}\) box is released on a \(23.5^\circ\) incline and accelerates down the incline at \(0.35 \, \text{m/s}^2\). Find the friction force impeding its motion. What is the coefficient of kinetic friction?
A rescue helicopter lifts a 79 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s2 and is lifted through a distance of 11 m.
An object weighs \( 432 \) \( \text{N} \) on the surface of Earth. At a height of \( 3R_{\text{Earth}} \) above Earth’s surface, what is its weight?
\(3.2 \times 10^3\,\text{km}\)
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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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