Step | Formula Derivation | Reasoning |
---|---|---|
1 | E_{\text{total}} = PE_{\text{max}} = KE_{\text{max}} | Total mechanical energy is constant, equal to maximum PE (at highest point) and maximum KE (at lowest point). |
2 | E_{\text{total}} = KE + PE | At any point, total mechanical energy is the sum of KE and PE. |
3 | PE = mgh | Potential energy at any point, where h is height above the lowest point. |
4 | mgh = \frac{1}{2} E_{\text{total}} | When KE equals PE, each is half the total mechanical energy. |
5 | h = L(1 – \cos(\theta)) | Height in terms of the pendulum length L and angle \theta. |
6 | L(1 – \cos(\theta)) = \frac{1}{2} L | Substitute h into the equation from Step 4, using E_{\text{total}} = mgL (max PE). |
7 | 1 – \cos(\theta) = \frac{1}{2} | Simplify the equation. |
8 | \cos(\theta) = \frac{1}{2} | Rearrange the equation to solve for \cos(\theta). |
9 | \theta \approx 60^\circ | Solve for \theta, knowing that \cos^{-1}\left(\frac{1}{2}\right) \approx 60^\circ. |
Therefore, the angle \theta at which the kinetic energy equals the potential energy is approximately \boxed{60^\circ} from the VERTICAL. From the horizontal this angle would be 90°-60° = 30° which is less than 45°. Note the questions ask for the angle from the horizontal.
Phy can also check your working. Just snap a picture!
How does the speed v1 of a block m reaching the bottom of slide 1 compare with v2, the speed of a block 2m reaching the end of slide 2? The blocks are released from the same height.
A pendulum bob of mass m on a cord of length L is pulled sideways until the cord makes an angle \theta with the vertical. The change in potential energy of the bob during the displacement is:
A satellite in circular orbit around the Earth moves at constant speed. This orbit is maintained by the force of gravity between the Earth and the satellite, yet no work is done on the satellite. How is this possible?
One end of a spring is attached to a solid wall while the other end just reaches to the edge of a horizontal, frictionless tabletop, which is a distance h above the floor. A block of mass M is placed against the end of the spring and pushed toward the wall until the spring has been compressed a distance x . The block is released and strikes the floor a horizontal distance D from the edge of the table. Air resistance is negligible.
Derive an expressions for the following quantities only in terms of M, x, D, h, and any constants.
A simple pendulum consists of a bob of mass 1.8 kg attached to a string of length 2.3 m. The pendulum is held at an angle of 30° from the vertical by a light horizontal string attached to a wall, as shown above.
By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
Kinematics | Forces |
---|---|
\Delta x = v_i t + \frac{1}{2} at^2 | F = ma |
v = v_i + at | F_g = \frac{G m_1m_2}{r^2} |
a = \frac{\Delta v}{\Delta t} | f = \mu N |
R = \frac{v_i^2 \sin(2\theta)}{g} |
Circular Motion | Energy |
---|---|
F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
a_c = \frac{v^2}{r} | PE = mgh |
KE_i + PE_i = KE_f + PE_f |
Momentum | Torque and Rotations |
---|---|
p = m v | \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 |
---|
F = -k x |
T = 2\pi \sqrt{\frac{l}{g}} |
T = 2\pi \sqrt{\frac{m}{k}} |
Constant | Description |
---|---|
g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
\mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
k | Spring constant, in \text{N/m} |
M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
Variable | SI Unit |
---|---|
s (Displacement) | \text{meters (m)} |
v (Velocity) | \text{meters per second (m/s)} |
a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
t (Time) | \text{seconds (s)} |
m (Mass) | \text{kilograms (kg)} |
Variable | Derived SI Unit |
---|---|
F (Force) | \text{newtons (N)} |
E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
P (Power) | \text{watts (W)} |
p (Momentum) | \text{kilogram meters per second (kgm/s)} |
\omega (Angular Velocity) | \text{radians per second (rad/s)} |
\tau (Torque) | \text{newton meters (Nm)} |
I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |
f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \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}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|
Pico- | p | 10^{-12} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
The most advanced version of Phy. Currently 50% off, for early supporters.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
A quick explanation
UBQ credits are specifically used to grade your FRQs and GQs.
You can still view questions and see answers without credits.
Submitting an answer counts as 1 attempt.
Seeing answer or explanation counts as a failed attempt.
Lastly, check your average score, across every attempt, in the top left.
MCQs are 1 point each. GQs are 1 point. FRQs will state points for each part.
Phy can give partial credit for GQs & FRQs.
Phy sees everything.
It customizes responses, explanations, and feedback based on what you struggle with. Try your best on every question!
Understand you mistakes quicker.
For GQs and FRQs, Phy provides brief feedback as to how you can improve your answer.
Aim to increase your understadning and average score with every attempt!
10 Free Credits To Get You Started
*Phy Pro members get unlimited credits
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.