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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] m = 110 \, \text{kg} [/katex] | Total mass of the person and bicycle is the sum of their individual masses. |
| 2 | [katex] d = 7 \, \text{m} [/katex] | Distance rolled down the hill before stopping. |
| 3 | [katex] \theta = 25^\circ [/katex] | Angle of inclination of the hill. |
| 4 | [katex] g = 9.8 \, \text{m/s}^2 [/katex] | Acceleration due to gravity. |
| 5 | [katex] \mu_k = 0.65 [/katex] | Coefficient of kinetic friction. |
| 6 | [katex] \Delta PE = mgh = mgd \sin(\theta) [/katex] | Calculate potential energy lost over the 7 m descent using the height change formula [katex] h = d \sin(\theta) [/katex]. |
| 7 | [katex] h = 7 \sin(25^\circ) [/katex] | Calculate the vertical distance descended. |
| 8 | [katex] \Delta PE = 110 \times 9.8 \times 7 \sin(25^\circ) [/katex] | Substitute known values into the potential energy formula. |
| 9 | [katex] \Delta PE \approx 2793.24 \, \text{J} [/katex] | Amount of energy needed to be transformed into friction, to bring the bike to a complete stop. |
| 9.5 | Calculate the maximum possible energy that friction can generate. | Use the [katex] W = Fd [/katex] as shown below |
| 10 | [katex] f_k = \mu_k (mg \cos(\theta)) [/katex] | Calculate the force of kinetic friction acting on the bicycle and rider while moving down the slope. |
| 11 | [katex] f_k = 0.65 \times (110 \times 9.8 \times \cos(25^\circ)) [/katex] | Substitute into the force of friction formula. |
| 12 | [katex] f_k \approx 627.89 \, \text{N} [/katex] | Calculating the force using approximate value of cosine function. |
| 13 | [katex] W = f_k \times d [/katex] | Work done by friction is the product of the force of friction and the distance over which it acts. |
| 14 | [katex] W = 627.89 \times 7 [/katex] | Substitute the values into the work formula. |
| 15 | [katex] W \approx 4395.23 \, \text{J} [/katex] | Calculating the total work done by friction. |
| 16 | Conclusion | The work done by friction exceeds the energy needed to bring the bike to a complete halt. Only 2793.24 J of frictional energy is transformed from potential energy to bring to bike to rest. |
Just ask: "Help me solve this problem."
A pendulum bob of mass m on a cord of length L is pulled sideways until the cord makes an angle [katex] \theta [/katex] with the vertical. The change in potential energy of the bob during the displacement is:
In which one of the following circumstances does the principle of conservation of mechanical energy apply, even though a nonconservative force acts on the moving object?
A car accelerates uniformly from rest to [katex] 29.4 [/katex] m/s in [katex] 6.93 [/katex] s along a level stretch of road. Ignoring friction, determine the average power in both watts and horsepower ([katex] 1 \text{ horsepower} = 745.7 \text{ Watts} [/katex]) required to accelerate the car if:
A 25 g steel ball is attached to the top of a 24-cm-diameter vertical wheel. Starting from rest, the wheel accelerates at [katex] 470 \, \frac{rad}{s^2}[/katex]. The ball is released after [katex]\frac{3}{4} [/katex] of a revolution. How high does it go above the center of the wheel?
Using only work and energy, find the velocity of the masses after they have traveled 0.8 meters. Refer to the image above.
2793.24 Joules
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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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