| 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 \(25 \, \text{g}\) steel ball is attached to the top of a \(24 \, \text{cm}\)-diameter vertical wheel of negligible mass. Starting from rest, the wheel accelerates at \(470 \, \text{rad/s}^2\). The ball is released after \(\frac{3}{4}\) of a revolution. How high does it go above the center of the wheel?
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 undergoing simple harmonic motion has a maximum displacement of \(6.2\) \(\text{m}\) at \(t = 0.0\) \(\text{s}\). If the angular frequency of oscillation is \(1.6\) \(\text{rad/s}\), what is the object’s displacement when \(t = 3.5\) \(\text{s}\)?
A bullet at speed \( v_0 \) trikes and embeds itself in a block of wood which is suspended by a string, causing the bullet and block to rise to a maximum height h. Which of the following statements is true?
A block is initially at rest on top of an inclined ramp that makes an angle \( \theta_0 \) with the horizontal. The distance measured along the base of the ramp is \( D \). After the block is released from rest, it slides down the frictionless ramp and then continues onto a rough horizontal surface until it finally comes to rest at the position \( x = 4D \) measured from the base of the ramp. The coefficient of kinetic friction between the block and the rough horizontal surface is \( \mu_k \).
You are working out on a rowing machine. Each time you pull the rowing bar toward you, it moves a distance of \(1.25 \, \text{m}\) in a time of \(0.84 \, \text{s}\). The readout on the display indicates that the average power you are producing is \(76 \, \text{W}\). What is the magnitude of the force that you exert on the handle?
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.
While traveling in its elliptical orbit around the Sun, Mars gains speed during the part of the orbit where it is getting closer to the Sun. Which of the following can be used to explain this gain in speed?
A ski tow carries people to the top of a nearby mountain. It operates on a slope of angle \( 15.7^\circ \) of length \( 260 \) \( \text{m} \). The rope moves at a speed of \( 13.0 \) \( \text{km/h} \) and provides power for \( 54 \) riders at one time, with an average mass per rider of \( 67.0 \) \( \text{kg} \).
A \(0.5 \, \text{kg}\) cart, on a frictionless \(2 \, \text{m}\) long table, is being pulled by a \(0.1 \, \text{kg}\) mass connected by a string and hanging over a pulley. The system is released from rest. After the hanging mass falls \(0.5 \, \text{m}\), calculate the speed of the cart on the table. Use ONLY forces and energy.
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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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