| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | F = ma | The net force acting on each vehicle due to friction is equal to the mass of the vehicle times its acceleration (Newton’s Second Law). |
| 2 | a = \frac{F}{m} | Acceleration a can be expressed as the force divided by the mass. Given that F is the same for both, the acceleration is inversely proportional to the mass. |
| 3 | v_f = v_i + at | The final velocity v_f of each vehicle can be found using the kinematic equation where v_i is the initial velocity and t is the time. |
| 4 | v_f = v – \frac{F}{m}t | Substituting for a in the kinematic equation. Note that different masses will lead to different decelerations if F is constant. |
| 5 | \Delta s = v_it + \frac{1}{2}at^2 | The distance \Delta s traveled under acceleration (deceleration in this case) is found using this formula, where the initial speed v_i is v, and the acceleration a is known. |
| 6 | D = vt – \frac{F}{2m}t^2 | Substituting -a (since it’s deceleration) and simplifying, configuring the equation in terms of t. |
| 7 | Solve for t | Solve the quadratic equation for t to find the time each vehicle takes to travel the distance D . Each vehicle, having different mass, will have a different t . |
| 8 | \text{Work done, } W = F \times D | The work done by the friction force over the distance D is the product of the force and the distance, which is the same for both since both F and D are constant. |
| 9 | P = \frac{W}{t} | Power P is defined as work done per unit time. Given the same work done but different times, power will be different. |
| 10 | KE = \frac{1}{2}mv^2_f | The kinetic energy at any point in time can be calculated using the mass and the velocity at that time. Since both mass and v_f differ for the two vehicles, kinetic energy will also differ. |
Correct answers based on these steps:
– (c) The work done on both vehicles is the same – since they are subject to the same force over the same distance. Other statements are false due to different masses and resulting differences in acceleration, velocity, time, and power.
Phy can also check your working. Just snap a picture!
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?
A boulder is raised above the ground so that its potential energy is 550 J. Then it is dropped. Assuming 92 J of energy was lost to air resistance, what is the kinetic energy of the boulder just before it hits the ground?
An object of mass 2 kg is thrown vertically downwards with an initial kinetic energy of 100 J. What is the distance fallen by the object at the instant when its kinetic energy has doubled?
A skier with a mass of 58 kg glides up a snowy incline that forms an angle of 28 degrees with the horizontal. The skier initially moves at a speed of 7.2 m/s. After traveling a distance of 2.3 meters up the slope, the skier’s speed reduces to 3.8 m/s.
An object with a mass m = 80 g is attached to a spring with a force constant k = 25 N/m. The spring is stretched 52.0 cm and released from rest. If it is oscillating on a horizontal frictionless surface, determine the velocity of the mass when it is halfway to the equilibrium position.
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} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
The most advanced version of Phy. Currently 50% off, for early supporters.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
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.
