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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 bullet of mass 0.0500 kg traveling at 50.0 m/s is fired horizontally into a wooden block suspended from a long rope. The mass of the wooden block is 0.300 kg and it is initially at rest. The collision is completely inelastic and after impact the bullet+ wooden block move together until the center of mass of the system rises a vertical distance h above its initial position.
A particle of mass m slides down a fixed, frictionless sphere of radius R, starting from rest at the top.
In terms of m, g, R, and O, determine each of the following for the particle while it is sliding on the sphere.
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 crane’s trolley at point \( P \) moves for a few seconds to the right with constant acceleration, and the \( 870 \, \text{kg} \) load hangs on a light cable at a \( 5^\circ \) angle to the vertical as shown. What is the acceleration of the trolley and load?
Three students are pulling on a bag of skittles. Each is pulling with a horizontal force. If student 1 pulls Eastward with [katex]170 \, \text{N}[/katex], student 2 pulls Southward with [katex]100 \, \text{N}[/katex] and student 3 pulls with [katex]200 \, \text{N}[/katex] at an angle of [katex]20^\circ [/katex] west of north, what is the net force caused by the three students on the bag of skittles?
[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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