| 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. |
A Major Upgrade To Phy Is Coming Soon — Stay Tuned
We'll help clarify entire units in one hour or less — guaranteed.
A self paced course with videos, problems sets, and everything you need to get a 5. Trusted by over 15k students and over 200 schools.
A baseball is thrown vertically into the air with a velocity \( v \), and reaches a maximum height \( h \). At what height was the baseball moving with one-half its original velocity? Assume air resistance is negligible.
Why do you need to “pump” your legs when you begin swinging on a park swing?
An airplane of weight \( W \) is flying horizontally with constant velocity. The total forward thrust of the engines is \( 3W \). What is the magnitude of the force of air on the plane in terms of \( W \)?
A \(0.5 \, \text{kg}\) cart, on a frictionless \(2 \, \text{m}\) long table, is being pulled by a \(0.1 \, \text{kg}\) mass connected by a string and hanging over a pulley. The system is released from rest. After the hanging mass falls \(0.5 \, \text{m}\), calculate the speed of the cart on the table. Use ONLY forces and energy.
When a basketball is dropped to the pavement, it bounces back up. Is a force needed to make it bounce back up? If so, what exerts the force?
For linear motion the term “inertia” refers to the same physical concept of
A student uses a pendulum to determine the acceleration due to gravity, \( g \). They measure the pendulum’s length \( L \) and its period \( T \). Which equation should they use to calculate \( g \)?
A linear spring of negligible mass requires a force of \( 18.0 \, \text{N} \) to cause its length to increase by \( 1.0 \, \text{cm} \). A sphere of mass \( 75.0 \, \text{g} \) is then attached to one end of the spring. The distance between the center of the sphere \( M \) and the other end \( P \) of the un-stretched spring is \( 25.0 \, \text{cm} \). Then the sphere begins rotating at constant speed in a horizontal circle around the center \( P \). The distance \( P \) and \( M \) increases to \( 26.5 \, \text{cm} \).
Late one morning, a mosquito collides with the windshield of a speeding truck. The force of the truck on the mosquito is ____ the force of the mosquito on the truck; the resulting acceleration of the mosquito is ____ the acceleration of the truck.
A crane’s trolley at point \( P \) moves for a few seconds to the right with constant acceleration, and the \( 870 \, \text{kg} \) load hangs on a light cable at a \( 5^\circ \) angle to the vertical as shown. What is the acceleration of the trolley and load?
By continuing you (1) agree to our Terms of Use and Terms of Sale 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 |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| 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] |
Metric Prefixes
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] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
One price to unlock most advanced version of Phy across all our tools.
per month
Billed Monthly. Cancel Anytime.
Try our free calculator to see what you need to get a 5 on the 2026 AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed
