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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] E_{\text{spring}} = \frac{1}{2} k x^2 [/katex] | The initial energy stored in the spring is given by the spring potential energy formula, where [katex] k [/katex] is the spring constant and [katex] x [/katex] is the compression. |
| 2 | [katex] E_{\text{spring}} = \frac{1}{2} \times 1800 \times 0.3^2 [/katex] | Substitute [katex] k = 1800 \, \text{N/m} [/katex] and [katex] x = 0.3 \, \text{m} [/katex] into the spring energy formula to calculate the initial energy. |
| 3 | [katex] E_{\text{spring}} = 81 \, \text{J} [/katex] | Calculate the total energy stored in the spring. |
| 4 | [katex] W_f = f_k \times d [/katex] | Calculate the work done by friction, where [katex] f_k [/katex] is the kinetic friction force and [katex] d [/katex] is the distance over which the force acts. |
| 5 | [katex] f_k = \mu_k \times m \times g [/katex] | Kinetic friction force is the coefficient of kinetic friction times the normal force. Here, [katex] \mu_k = 0.12 [/katex], [katex] m = 6 \, \text{kg} [/katex], and [katex] g = 9.8 \, \text{m/s}^2 [/katex]. On horizontal surface, normal force equals mg. |
| 6 | [katex] f_k = 0.12 \times 6 \times 9.8 = 7.056 \, \text{N}[/katex] | Calculate the kinetic friction force on the horizontal surface. |
| 7 | [katex] W_f = 7.056 \times 3.3 = 23.28 \, \text{J} [/katex] | Calculate the work done by friction over the 3 m horizontal surface and the distance it slides while the spring decompresses (.3 m) for a total distance of 3.3 m. |
| 8 | [katex] W_{f, \text{slope}} = f_{k, \text{slope}} \times d_{\text{slope}} [/katex] | Work done by friction on the slope, where [katex] f_{k, \text{slope}} [/katex] is the kinetic friction force on the slope and [katex] d_{\text{slope}} = 4.5 \, \text{m} [/katex]. |
| 9 | [katex] f_{k, \text{slope}} = \mu_{k, \text{slope}} \times m \times g \times \cos(\theta) [/katex] | On the slope, the normal force is reduced by the cosine of the slope angle [katex] \theta = 12^\circ [/katex]. |
| 10 | [katex] W_{f, \text{slope}} = \mu_{k, \text{slope}} \times m \times g \times \cos(\theta) \times d_{\text{slope}} [/katex] | Substitute equation from step 9 into the equation for work done by the friction of the slope (step 8). |
| 11 | [katex] W_{f, \text{slope}} = \mu_{k, \text{slope}} \times 258.818 \, \text{J}[/katex] | Plug in given values and simplify the equation for work done by the friction of the slope as much as possible. |
| 12 | [katex] PE_{\text{top}} = mgh = 5 \times 9.8 \times 4.5sin(12^\circ) \approx 55.01 \, \text{J}[/katex] | Now find the final potential energy at the top of the incline. |
| 13 | [katex] E_{\text{spring}} = W_f + W_{f, \text{slope}} + PE_{\text{top}} [/katex] | Finally add up all the energy types together. Use the law of conservation of energy which tells us the initial energy from the spring will turn into work done by friction on level ground, work done by friction on the slope, and the potential energy gained traveling up the slope. |
| 14 | [katex] 81 = 23.28 + (\mu_{k, \text{slope}} \times 258.818) + 55.01 [/katex] | Substitute all numbers found in the previous steps. |
| 15 | [katex] \boxed{\mu_{k, \text{slope}} \approx .01} [/katex] | Solve for the coefficient of kinetic friction on the slope. |
Just ask: "Help me solve this problem."
A spacecraft somewhere in between the earth and the moon experiences 0 net force acting on it. This is because the earth and the moon pull the spacecraft in equal but opposite directions. Find the distance D away from Earth, such that the spacecraft experiences zero net force. The distance between the Moon and Earth is ~3.844 x 108 m.
NOTE: You may need the mass of the earth and moon. You can find this in the formula table.
The box is sitting on the floor of an elevator. The elevator is accelerating upward. The magnitude of the normal force on the box is
A ball of mass \(m\) is released from rest at a distance \(h\) above a frictionless plane inclined at an angle of \(45^\circ\) to the horizontal as shown above. The ball bounces horizontally off the plane at point \(P_1\) with the same speed with which it struck the plane and strikes the plane again at point \(P_2\). In terms of \(g\) and \(h\), determine each of the following quantities:
An object has a mass of 10 kg. For each case below answer the questions and provide an example.
A block rests on a flat plane inclined at an angle of 30° with respect to the horizontal. What is the minimum coefficient of friction necessary to keep the block from sliding?
[katex] \mu_{k, \text{slope}} \approx .01 [/katex]
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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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