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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[a = \frac{F}{m}\] | Using Newton’s second law with \(F = 4.40\,\text{N}\) and \(m = 1.00\,\text{kg}\) gives the constant horizontal acceleration. |
| 2 | \[t_1 = \sqrt{\frac{2 \Delta x}{a}}\] | Time \(t_1\) for the puck to traverse \(\Delta x = 2.10\,\text{m}\) on the table with \(v_i = 0\) and constant acceleration \(a\). |
| 3 | \[v_x = a t_1\] | Horizontal speed at the table edge after accelerating for time \(t_1\). |
| 4 | \[t_2 = \sqrt{\frac{2 h}{g}}\] | Time of fall \(t_2\) from height \(h = 2.10\,\text{m}\) with initial vertical velocity zero under gravity \(g = 9.80\,\text{m/s}^2\). |
| 5 | \[\Delta x_2 = v_x t_2 + \frac{1}{2} a t_2^{2}\] | Horizontal distance while airborne: initial horizontal speed \(v_x\) for time \(t_2\) plus extra distance from constant thrust acceleration. |
| 6 | \[\Delta x_2 \approx 3.76\,\text{m}\] | Substituting \(a = 4.40\,\text{m/s}^2\), \(t_1 = \sqrt{\tfrac{2(2.10)}{4.4}} \approx 0.98\,\text{s}\), \(v_x \approx 4.30\,\text{m/s}\), and \(t_2 = \sqrt{\tfrac{2(2.10)}{9.8}} \approx 0.65\,\text{s}\) gives the horizontal distance from the edge. |
| 7 | \[\boxed{3.76\,\text{m}}\] | The puck lands \(3.76\,\text{m}\) from the base of the table. |
Just ask: "Help me solve this problem."
A block of weight \( W \) is pulled along a horizontal surface at constant speed by a force \( F \), which acts at an angle of \( \theta \) with the horizontal. The normal force exerted on the block by the surface has magnitude:
A \(1.5 \, \text{kg}\) object is located at a distance of \(1.7 \times 10^{6} \, \text{m}\) from the center of a larger object whose mass is \(7.4 \times 10^{22} \, \text{kg}\).
A toy car moves off the edge of a table that is \(1.25 \, \text{m}\) high. If the car lands \(0.40 \,\text{m}\) from the base of the table…
A pulley system consists of two blocks of mass \( 5 \) \( \text{kg} \) and \( 10 \) \( \text{kg} \), connected by a rope of negligible mass that passes over a pulley of radius \( 0.1 \) \( \text{m} \) and mass \( 2 \) \( \text{kg} \). The pulley is free to rotate about its axis. The system is released from rest, and the block of mass \( 10 \) \( \text{kg} \) starts to move downwards. Assume the pulley has a frictional force of \(5.7\) Newtons acting on the outer edge of the pulley.
A linear spring of negligible mass requires a force of \( 18.0 \, \text{N} \) to cause its length to increase by \( 1.0 \, \text{cm} \). A sphere of mass \( 75.0 \, \text{g} \) is then attached to one end of the spring. The distance between the center of the sphere \( M \) and the other end \( P \) of the un-stretched spring is \( 25.0 \, \text{cm} \). Then the sphere begins rotating at constant speed in a horizontal circle around the center \( P \). The distance \( P \) and \( M \) increases to \( 26.5 \, \text{cm} \).
\(3.76\,\text{m}\)
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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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