| 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. |
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A simple pendulum consists of a sphere tied to the end of a string of negligible mass. The sphere is pulled back until the string is horizontal and then released from rest. Assume the gravitational potential energy is zero when the sphere is at its lowest point.
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A \( 0.30 \text{-kg} \) mass is suspended on a spring. In equilibrium the mass stretches the spring \( 2.0 \) \( \text{cm} \) downward. The mass is then pulled an additional distance of \( 1.0 \) \( \text{cm} \) down and released from rest. Write down its equation of motion.
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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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