| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | KE_{\text{initial}} = \frac{1}{2} m v^2 | Calculate the initial kinetic energy of the vehicle using its speed before it halts. Here v = 12.3 \text{ m/s}. |
| 2 | W_{\text{friction}} = f_k \cdot d = \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d | The work done by friction, where \mu_k = 0.650 is the coefficient of kinetic friction, g = 9.8 \text{ m/s}^2 is acceleration due to gravity, \theta = 18^\circ, and d is the distance the vehicle slides. |
| 3 | W_{\text{gravity}} = m \cdot g \cdot \sin(\theta) \cdot d | The work done by gravity while the vehicle moves down the incline. |
| 4 | KE_{\text{initial}} = W_{\text{friction}} + W_{\text{gravity}} | By the work-energy principle, the initial kinetic energy is converted into work done against friction plus the work done by gravity. |
| 5 | \frac{1}{2} m v^2 = \mu_k mg\cos(\theta)d + mg\sin(\theta)d | Substitute expressions from steps 1, 2, and 3 into the work-energy equation. |
| 6 | d = \frac{\frac{1}{2} v^2}{\mu_k g \cos(\theta) + g \sin(\theta)} | Solve for d, distance the vehicle slides. Notice that mass m cancels out. |
| 7 | d = \frac{\frac{1}{2} (12.3)^2}{0.650 \times 9.8 \times \cos(18^\circ) + 9.8 \times \sin(18^\circ)} | Substitute numerical values for v, \mu_k, g, and \theta to find the value of d that represents the distance the vehicle slides until it stops. |
| 8 | d = 8.32 \,\text{m} | Calculated value. |
To solve part b, look at the equation derived in step 6 of part a. Notice that the angle and distance traveled up the incline are inversely proportional. This means the greater the angle \theta the shorter the distance traveled.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | d' = \frac{\frac{1}{2} v^2}{\mu_k g \cos(27^\circ) + g \sin(27^\circ)} | Recalculate the distance with the increased angle of 1.5 \times 18^\circ = 27^\circ. |
| 2 | d' = \frac{d}{\cos(27^\circ) + \tan(27^\circ) \cdot \sin(27^\circ)} | Using the previous formula of d, we express the new sliding distance d' in terms of the old distance d. |
| 3 | d' = 7.41 \, \text{meters} | At an angle of 27^\circ the vehicale would slide up 7.41 meters, which is less than the orginal dsitance of 8.32 meters. |
| 3 | \frac{d'}{d} | The ratio \frac{d'}{d} shows how much further the vehicle would slide relative to d. |
| 4 | \frac{7.41}{8.32} = .89 | Thus the new distance d' = .89d |
Part (c):
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | W_{\text{gravity,up}} = -m \cdot g \cdot \sin(\theta) \cdot d | The work done by gravity as the vehicle slides up the incline is negative since gravity opposes the motion. |
| 2 | \frac{1}{2} m v^2 = \mu_k m g \cos(\theta) d – m g \sin(\theta) d | Work-energy principle applied while moving up. The kinetic energy has to overcome both friction and an upward gravity force. |
| 3 | d_{\text{up}} = \frac{\frac{1}{2} v^2}{\mu_k g \cos(\theta) – g \sin(\theta)} | Solve for the distance the vehicle would slide up the incline. |
| 4 | d_{\text{up}} < d | The distance d_{\text{up}} will be lesser than d since gravity now acts against the motion, reducing the sliding distance relative to sliding down. |
Phy can also check your working. Just snap a picture!
A 75.0kg log floats downstream with a speed of 1.80 m/s. Eight frogs hop onto the log in a series of perfectly inelastic collisions. If each frog has a mass of 0.30 kg and an upstream speed of 1.3 m/s, what is the change in kinetic energy for this system?
A 1.5 kg block is pushed to the right with just enough force to get it to move. The block is pushed for five seconds with this constant force, then the force is released and the block slides to a stop. If the coefficient of kinetic friction is 0.300 and the coefficient of static friction is 0.400. Calculate the amount of time that passes from when the force is applied to when the block stops.
When the speed of a rear-wheel-drive car is increasing on a horizontal road, what is the direction of the frictional force on the tires?
Block 2 initially is at rest. Block 1 travels towards block 2 and collides with Block 2 as shown above. Find the final velocities of both blocks assuming the collision is elastic.
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?
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.
