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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 small block moving with a constant speed v collides inelastically with a block M attached to one end of a spring k. The other end of the spring is connected to a stationary wall. Ignore friction between the blocks and the surface.
A kickball is rolled by the pitcher at a speed of 10 m/s and it is kicked by another student. The kickball deforms a little during the kick, and then rebounds with a velocity of 15 m/s as its shape restores to a perfect sphere. Select all that must be true about the kickball and the kicking foot system.
A person is trying to judge whether a picture (mass = 1.42 kg) is properly positioned by temporarily pressing it against a wall. The pressing force is perpendicular to the wall. The coefficient of static friction between the picture and the wall is 0.62. What is the minimum amount of pressing force that must be used?
A 2,000 kg car collides with a stationary 1,000 kg car. Afterwards, they slide 6 m before coming to a stop. The coefficient of friction between the tires and the road is 0.7. Find the initial velocity of the 2,000 kg car before the collision?
Two balls are thrown off a building with the same speed, one straight up and one at a 45° angle. Which statement is true if air resistance can be ignored?
[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] | 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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