| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | \(F_{\text{parallel}} = mg\sin(\theta)\) | Force parallel to the incline, where \(m\) is mass, \(g\) is acceleration due to gravity, and \(\theta\) is the incline angle. |
| 2 | \(F_{\text{normal}} = mg\cos(\theta)\) | Normal force, perpendicular to the incline. |
| 3 | \(F_{\text{friction}} = \mu_k F_{\text{normal}}\) | Frictional force, where \(\mu_k\) is the coefficient of kinetic friction. |
| 4 | \(F_{\text{friction}} = 0.250 \times mg\cos(25^\circ)\) | Substitute values for \(\mu_k\) and \(F_{\text{normal}}\). |
| 5 | \(F_{\text{pull}} = F_{\text{parallel}} + F_{\text{friction}}\) | Total force to pull the crate includes parallel force and frictional force. |
| 6 | \(F_{\text{pull}} = mg\sin(25^\circ) + 0.250 \times mg\cos(25^\circ)\) | Combine parallel and frictional forces. |
| 7 | \(W_{\text{output}} = F_{\text{parallel}} \times d\) | Work output, where \(F_{\text{parallel}}\) is the force parallel to the incline and \(d\) is the distance. |
| 8 | \(W_{\text{output}} = mg\sin(25^\circ) \times 2.5\) | Work done against gravity. |
| 9 | \(W_{\text{input}} = F_{\text{pull}} \times d\) | Work input, force to pull the crate. |
| 10 | \(W_{\text{input}} = \left( mg\sin(25^\circ) + 0.250 \times mg\cos(25^\circ) \right) \times 2.5\) | Work done to pull the crate including overcoming friction. |
| 11 | \(\text{Efficiency} = \frac{W_{\text{output}}}{W_{\text{input}}} \times 100%\) | Efficiency formula. |
| 12 | \(\text{Efficiency} = \frac{mg\sin(25^\circ) \times 2.5}{\left( mg\sin(25^\circ) + 0.250 \times mg\cos(25^\circ) \right) \times 2.5} \times 100%\) | Substitute \(W_{\text{output}}\) and \(W_{\text{input}}\). |
| 13 | \(\text{Efficiency} = \frac{\sin(25^\circ)}{\sin(25^\circ) + 0.250 \times \cos(25^\circ)} \times 100%\) | Simplify by canceling \(mg\) and \(d\). |
| 14 | \(\text{Efficiency} = 65.10\%\) | Calculated efficiency. |
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A pendulum consists of a ball of mass \( m \) suspended at the end of a massless cord of length \( L \). The pendulum is drawn aside through an angle of \( 60^\circ \) with the vertical and released. At the low point of its swing, the speed of the pendulum ball is
A spring stretches \( 8.0 \) \( \text{cm} \) when a \( 13 \) \( \text{N} \) force is applied. How far does it stretch when a \( 26 \) \( \text{N} \) force is applied?
An object with a mass \(m = 80 \, \text{g}\) is attached to a spring with a force constant \(k = 25 \, \text{N/m}\). The spring is stretched \(52.0 \, \text{cm}\) and released from rest. If it is oscillating on a horizontal frictionless surface, determine the velocity of the mass when it is halfway to the equilibrium position.
A \(2,000 \, \text{kg}\) car collides with a stationary \(1,000 \, \text{kg}\) car. Afterwards, they slide \(6 \, \text{m}\) before coming to a stop. The coefficient of friction between the tires and the road is \(0.7\). Find the initial velocity of the \(2,000 \, \text{kg}\) car before the collision?
A \(2 \, \text{kg}\) model rocket is launched with a thrust force of \(275 \, \text{N}\) and reaches a height of \(90 \, \text{m}\), at which point the thrust cuts out, but the rocket continues moving at \(150 \, \text{m/s}\). What is the average air resistance force acting on the rocket during its ascent?
A man weighing \( 700 \) \( \text{N} \) and a woman weighing \( 400 \) \( \text{N} \) have the same momentum. What is the ratio of the man’s kinetic energy \( K_m \) to that of the woman \( K_w \)?
Water balloons are tossed from the roof of a building, all with the same speed but with different launch angles. Which one has the highest speed when it hits the ground? Ignore air resistance. Without using equations, explain your answer.
A vehicle is moving at a speed of 12.3 m/s on a decline when the brakes of all four wheels are fully applied, causing them to lock. The slope of the decline forms an angle of 18.0 degrees with the horizontal plane. Given that the coefficient of kinetic friction between the tires and the road surface is 0.650.
A girl throws a stone from a bridge. Consider the following ways she might throw the stone. The speed of the stone as it leaves her hand is the same in each case.
Case A: Thrown straight up.
Case B: Thrown straight down.
Case C: Thrown out at an angle of 45° above horizontal.
Case D: Thrown straight out horizontally.
In which case will the speed of the stone be greatest when it hits the water below if there is no significant air resistance, assuming equal initial speeds?
A 84.4 kg climber is scaling the vertical wall. His safety rope is made of a material that behaves like a spring that has a spring constant of 1.34 x 103 N/m. He accidentally slips and falls 0.627 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?
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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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