| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | KE_{\text{peak}} = \frac{1}{2} m v^2 | Calculate the kinetic energy at the peak height. Kinetic energy (KE) is given by the equation \frac{1}{2} m v^2, where m is mass (2 kg) and v is velocity (150 m/s). |
| 2 | PE_{\text{peak}} = mgh | Calculate the potential energy at the peak height. Potential energy (PE) is given by mgh where m is mass (2 kg), g is the acceleration due to gravity (approximately 9.8 \, \text{m/s}^2) and h is the height (90 m). |
| 3 | W_{\text{thrust}} = F_{\text{thrust}} \times h | Calculate the work done by the thrust force. Work (W) is force (F) times the displacement (h). Here, F_{\text{thrust}} is 275 N. |
| 4 | KE_{\text{peak}} = \frac{1}{2} \times 2 \times 150^2 = 22500 \, \text{J} | Substitute the values to find KE_{\text{peak}} . |
| 5 | PE_{\text{peak}} = 2 \times 9.8 \times 90 = 1764 \, \text{J} | Substitute the values to find PE_{\text{peak}} . |
| 6 | W_{\text{thrust}} = 275 \times 90 = 24750 \, \text{J} | Substitute the values to find W_{\text{thrust}} . |
| 7 | W_{\text{thrust}} = KE_{\text{peak}} + PE_{\text{peak}} + W_{AR} | Use the conservation of energy. In this case the initial work done by the rocket thrust transforms into kinetic energy, potential energy, and work done by air resistance W_{AR} . |
| 8 | W_{AR} = 24750 – 22500 – 1764 | Solving for the work done by air resistance gives us W_{AR} = 486 \, \text{J} . |
| 9 | W_{AR} = Fh | Divide the work done by air resistance by the height to find the average air resistance force. |
| 10 | W_{AR} = \frac{486}{90} \approx 5.4 \, \text{N} | F_{\text{air}} \approx \boxed{5.4} \, \text{N} is the average air resistance force acting against the rocket during ascent. |
ALTERNATE EXPLANATION – Using Forces
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | F_{\text{net}} = ma | Newton’s second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. |
| 2 | F_{\text{net}} = F_{\text{thrust}} – F_{\text{gravity}} – F_{\text{drag}} | The net force acting on the rocket includes the thrust force, the gravitational force, and the drag force (air resistance), where the thrust and drag act upwards, and gravity acts downwards. |
| 3 | F_{\text{gravity}} = mg | The gravitational force is calculated as the product of the mass and the acceleration due to gravity g \approx 9.81 \, \text{m/s}^2 . |
| 4 | F_{\text{gravity}} = 2 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 19.62 \, \text{N} | Substitute the values of the mass and gravitational acceleration to find the gravitational force. |
| 5 | a = \frac{v^2 – v_0^2}{2h} | The kinematic equation relates the rocket’s change in velocity to acceleration and distance. Here, v_0 = 0 \, \text{m/s} (initial velocity), v = 150 \, \text{m/s} (final velocity at the peak), and h = 90 \, \text{m} (height). |
| 6 | a = \frac{(150 \, \text{m/s})^2}{2 \times 90 \, \text{m}} | Substituting the values into the kinematic equation to calculate acceleration. |
| 7 | a = 125 \, \text{m/s}^2 | Calculation of the acceleration using the given values. |
| 8 | F_{\text{net}} = 2 \, \text{kg} \times 125 \, \text{m/s}^2 = 250 \, \text{N} | Calculate the net force acting on the rocket using Newton’s second law with the rocket’s mass and the acceleration. |
| 9 | 250 \, \text{N} = 275 \, \text{N} – 19.62 \, \text{N} – F_{\text{drag}} | Solve for the drag force using the equation in step 2. |
| 10 | F_{\text{drag}} = 275 \, \text{N} – 19.62 \, \text{N} – 250 \, \text{N} = 5.38 \, \text{N} | Substitute the values of the thrust, gravitational, and net forces to find the drag force. |
| 11 | F_{\text{drag}} \approx 5.38 \, \text{N} | This is the average air resistance force acting on the rocket during its ascent. |
Phy can also check your working. Just snap a picture!
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 car traveling to the right with a speed v brakes to a stop in a distance d. What is the work done on the car by the frictional force F? (Assume that the frictional force is constant)
A simple pendulum consists of a sphere tied to the end of a string of negligible mass. The sphere is pulled back until the string is horizontal and then released from rest. Assume the gravitational potential energy is zero when the sphere is at its lowest point.
What angle will the string make with the horizontal when the kinetic energy and the potential energy of the sphere-Earth system are equal?
A vehicle is moving at a speed of 12.3 m/s on a decline when the brakes of all four wheels are fully applied, causing them to lock. The slope of the decline forms an angle of 18.0 degrees with the horizontal plane. Given that the coefficient of kinetic friction between the tires and the road surface is 0.650.
A girl throws a stone from a bridge. Consider the following ways she might throw the stone. The speed of the stone as it leaves her hand is the same in each case.
Case A: Thrown straight up.
Case B: Thrown straight down.
Case C: Thrown out at an angle of 45° above horizontal.
Case D: Thrown straight out horizontally.
In which case will the speed of the stone be greatest when it hits the water below if there is no significant air resistance, assuming equal initial speeds?
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.
