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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 1.0-kg object is moving with a velocity of 6.0 m/s to the right. It collides and sticks to a 2.0-kg object moving with a velocity of 3.0 m/s in the same direction. How much kinetic energy was lost in the collision?
How does the speed \(v_1\) of a block \(m\) reaching the bottom of slide 1 compare with \(v_2\), the speed of a block \(2m\) reaching the end of slide 2? The blocks are released from the same height.
A net force of \( 8.0 \) \( \text{N} \) accelerates a \( 4.0 \) \( \text{kg} \) body from rest to a speed of \( 5.0 \) \( \text{m s}^{-1} \). Which of the following is equal to the work done by the force?
A block of mass 3.0 kg is hung from a spring, causing it to stretch 12 cm at equilibrium. The 3.0 kg block is then taken off and the spring returns to its original height. Now a 4.0 kg block is placed on the spring and released from rest. How far will the 4.0 kg block fall before its direction is reversed?
How does the time t1 of a block m reaching the bottom of slide 1 compare with t2, the time taken block of mass 2m to reach the end of slide 2 that’s curved? The blocks are released from the same height.
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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