| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta x = \frac{1}{2} a t^2\] | Use constant–acceleration kinematics with initial velocity \(v_i = 0\). |
| 2 | \[a = \frac{2\Delta x}{t^2}\] | Algebraically isolate \(a\). |
| 3 | \[a = \frac{2(10)}{(4.5)^2} = 0.99\,\text{m/s}^2\] | Insert \(\Delta x = 10\,\text{m}\) and \(t = 4.5\,\text{s}\) to find \(a \approx 0.99\,\text{m/s}^2\). |
| 4 | \[F_{1x}=100\cos20^\circ,\;F_{2x}=-400\cos40^\circ,\;F_{3x}=500\cos10^\circ\] | Resolve each force into horizontal components; left is negative, right positive. |
| 5 | \[F_{1x}=93.97\,\text{N},\;F_{2x}=-306.6\,\text{N},\;F_{3x}=492.4\,\text{N}\] | Evaluate the trigonometric products. |
| 6 | \[F_{x,\text{applied}} = 279.8\,\text{N}\] | Sum \(F_{1x}+F_{2x}+F_{3x}\) to get the net applied horizontal force. |
| 7 | \[F_{1y}=+100\sin20^\circ,\;F_{2y}=-400\sin40^\circ,\;F_{3y}=-500\sin10^\circ\] | Resolve forces into vertical components; upward is positive. |
| 8 | \[F_{1y}=+34.2\,\text{N},\;F_{2y}=-257.1\,\text{N},\;F_{3y}=-86.8\,\text{N}\] | Compute the numeric values. |
| 9 | \[F_{y,\text{applied}} = -309.7\,\text{N}\] | Add the vertical components to find a downward net external load of \(309.7\,\text{N}\). |
| 10 | \[N = mg + 309.7\,\text{N}\] | Vertical equilibrium requires \(N + F_{1y} = mg + |F_{2y}| + |F_{3y}|\); simplify to this expression for the normal force. |
| 11 | \[f_k = \mu_k N = 0.2\,(mg + 309.7)\] | Kinetic friction opposes motion with magnitude \(\mu_k N\). |
| 12 | \[F_{\text{net}} = F_{x,\text{applied}} – f_k\] | Net horizontal force equals applied force minus friction (leftward). |
| 13 | \[ma = 279.8 – 0.2(mg + 309.7)\] | Apply Newton’s second law \(\sum F_x = ma\). |
| 14 | \[m(a + 1.96) = 217.8\] | Use \(g = 9.8\,\text{m/s}^2\) and simplify: \(0.2g = 1.96\,\text{N/kg}\). |
| 15 | \[m = \frac{217.8}{a + 1.96}\] | Isolate the unknown mass. |
| 16 | \[m = \frac{217.8}{0.99 + 1.96} = 73.9\,\text{kg}\] | Insert \(a \approx 0.99\,\text{m/s}^2\) to calculate \(m\). |
| 17 | \[\boxed{m \approx 7.4 \times 10^{1}\,\text{kg}}\] | Express the mass to two significant figures. |
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A projectile is launched at \( 25 \) \( \text{m/s} \) at an angle of \( 45^\circ \). It lands on a slope \( 5 \) \( \text{m} \) below the launch height. On landing, it rebounds vertically with \( 80\% \) of its speed and falls straight down from there. Find the total time from launch to final impact at the base of the slope.
A horizontal uniform meter stick of mass 0.2 kg is supported at its midpoint by a pivot point. A mass of 0.1 kg is attached to the left end of the meter stick, and another mass of 0.15 kg is attached to the right end of the meter stick. The meter stick is free to rotate in the horizontal plane around the pivot point. What is the tension in the string supporting the left end of the meter stick?
A cart is initially moving at 0.5 m/s along a track. The cart comes to rest after traveling 1 m. The experiment is repeated on the same track, but now the cart is initially moving at 1 m/s. How far does the cart travel before coming to rest?
Two masses, \( m_1 \) and \( m_2 \), are connected by a cord and arranged as shown in the diagram, with \( m_1 \) sliding along a frictionless surface and \( m_2 \) hanging from a light, frictionless pulley. What would be the mass of the falling mass, \( m_2 \), if both the sliding mass, \( m_1 \), and the tension, \( T \), in the cord were known?
A boat is rowed directly upriver at a speed of \(2.5 \, \text{m/s}\) relative to the water. Viewers on the shore find that it is moving at only \(0.5 \, \text{m/s}\) relative to the shore. What is the speed of the river? Is it moving with or against the boat?
A box with a mass of \( M \) rests on a scale in an elevator that is moving downwards. The elevator slows with an acceleration of \( \dfrac{g}{4} \). Which of the following will give the reading of the scale?
Which statement is true about the distances the two objects have traveled at time \( t_f \)?
You stand at the edge of a vertical cliff and throws a stone vertically upwards. The stone leaves your hand with a speed v = 8.0 m/s. The time between the stone leaving your hand and hitting the sea is 3.0 s. Assume air resistance is negligible. Calculate:
An adult exerts a horizontal force on a swing that is suspended by a rope of length \( L \), holding it at an angle \( \theta \) with the vertical. The child in the swing has a weight \( W \) and dimensions that are negligible. In terms of \( W \) and \( \theta \), determine:
The figure above shows a cart of mass \( M \) accelerating to the right with acceleration \( a \). A block of mass \( m \) is pressed against the cart’s front vertical surface and is held there only by friction. The coefficient of friction between the block and the cart is \( \mu \). What is the minimum acceleration \( a \) of the cart such that the block will not fall?
\(74\,\text{kg}\)
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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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