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| Derivation/Formula | Reasoning |
|---|---|
| \[\text{Forces:}\; N,\; mg,\; f_s\] | The diagram contains the normal reaction \(N\) acting perpendicular to the road surface, the weight \(mg\) acting vertically downward, and the static–friction force \(f_s\) acting up–slope while the rider is at rest on the hill. |
| \[mg\sin\theta,\; mg\cos\theta\] | The weight is resolved into components parallel (\(mg\sin\theta\), down–slope) and perpendicular (\(mg\cos\theta\)) to the incline of angle \(\theta=30^{\circ}\). |
| \[f_s\le \mu_sN\] | Static friction adjusts up to its maximum value \(\mu_sN\;(\mu_s=0.85)\) to keep the bicycle from sliding while the rider pauses. |
| Derivation/Formula | Reasoning |
|---|---|
| \[m=90\,\text{kg}+12\,\text{kg}=102\,\text{kg}\] | Total mass is the sum of rider and bicycle. |
| \[h=\Delta x_1\sin\theta=9\,(0.5)=4.5\,\text{m}\] | The vertical drop after rolling \(\Delta x_1=9\,\text{m}\) down a \(30^{\circ}\) incline is \(h=\Delta x_1\sin\theta\). |
| \[v_i=\sqrt{2gh}=\sqrt{2(9.8)(4.5)}\] | Conservation of energy (no non-conservative work before braking) gives the speed \(v_i\) at the instant the wheels lock. |
| \[v_i\approx9.4\,\text{m\,s}^{-1}\] | Numeric evaluation of the previous expression. |
| \[F_f=\mu_kN=\mu_kmg\cos\theta\] | Once the wheels are locked, kinetic friction of magnitude \(F_f\) opposes the motion (\(\mu_k=0.7\)). |
| \[F_f=0.7(102)(9.8)(0.866)=6.06\times10^{2}\,\text{N}\] | Calculating the friction force with \(\cos30^{\circ}=0.866\). |
| \[\Delta x_2=\frac{\tfrac12v_i^{2}}{g(\mu_k\cos\theta-\sin\theta)}\] | Work–energy: net work \(mg\sin\theta\Delta x_2-F_f\Delta x_2=-\tfrac12mv_i^{2}\). Solving for stopping distance \(\Delta x_2\). |
| \[\Delta x_2\approx42.5\,\text{m}\] | Numeric substitution using \(\mu_k\cos\theta-\sin\theta\approx0.106\). |
| \[W_f=-F_f\Delta x_2\] | Work done by friction is negative because it opposes the displacement down the slope. |
| \[\boxed{W_f\approx-2.6\times10^{4}\,\text{J}}\] | Final numeric value of energy removed by kinetic friction to bring the bicycle to rest. |
| Derivation/Formula | Reasoning |
|---|---|
| \[F_{\text{up}}=mg\sin\theta+f\] | When climbing at constant speed the cyclist’s legs must generate an up-slope force equal to gravity’s component \(mg\sin\theta\) plus rolling/drag/friction forces \(f\). |
| \[F_{\text{down}}=mg\sin\theta-f\] | During descent gravity supplies \(mg\sin\theta\); only a portion is cancelled by friction or air drag, so the net driving force is \(mg\sin\theta-f\) acting without muscular effort. |
| \[F_{\text{up}}>F_{\text{down}}\] | Because \(f>0\), the force required from the cyclist on the climb is strictly larger than the net force aiding motion downhill, making ascent harder than descent. |
| \[W_{\text{climb}}=mg\,h\;>\;0,\quad W_{\text{down}}=-mg\,h\] | Energy must be supplied to increase gravitational potential when going up (positive work), whereas gravity returns that energy on the way down (negative work done by the rider), confirming the greater effort needed to ascend. |
Just ask: "Help me solve this problem."
A linear spring of force constant \( k \) is used in a physics lab experiment. A block of mass \( m \) is attached to the spring and the resulting frequency, \( f \), of the simple harmonic oscillations is measured. Blocks of various masses are used in different trials, and in each case, the corresponding frequency is measured and recorded. If \( f^{2} \) is plotted versus \( \frac{1}{m} \), the graph will be a straight line with slope
A person holds a book at rest a few feet above a table. The person then lowers the book at a slow constant speed and places it on the table. Which of the following accurately describes the change in the total mechanical energy of the Earth–book system?
A mechanic pushes a [katex]2500 \, \text{kg}[/katex] car from rest to a final speed [katex]v[/katex] by doing [katex]5.0 \times 10^3 \, \text{J}[/katex] of work on the car. Frictional effect between the car and the ground are negligible. What is the final speed of the car?
A snowboarder starts from rest and slides down a \(32^\circ\) incline that’s \(75 \, \text{m}\) long.
A typical \( 68 \text{-kg} \) person generates a steady mechanical power output of \( 120 \text{ W} \) at the pedals of a bicycle. Approximately how many Calories are “burned” (total metabolic energy expended) when the person rides a bicycle for \( 15 \text{ minutes} \)? A typical energy efficiency for the human body is \( 25\% \), which takes into account the release of thermal energy. Note (\( 1 \text{ Cal} = 4186 \text{ J} \)).
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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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