0 attempts
0% avg
UBQ Credits
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[a_{\text{level}} = \mu g \quad \Longrightarrow \quad \mu = \frac{3.80}{g}\] | On a level road the maximum deceleration is provided entirely by static friction. Since the car decelerates at \(-3.80 \, \text{m/s}^2\) (in magnitude \(3.80\, \text{m/s}^2\)), we can write \(a_{\text{level}} = \mu g\) and solve for the friction coefficient \(\mu\). |
| 2 | \[N = mg\cos(9.3^\circ)\] | On an incline the normal force is reduced to \(mg\cos(9.3^\circ)\); this reduction affects the maximum static friction available. |
| 3 | \[f_{\text{max}} = \mu mg\cos(9.3^\circ)\] | The maximum frictional force that can be provided on the incline is given by \(\mu N\), which is \(\mu mg\cos(9.3^\circ)\). |
| 4 | \[a_{\text{friction}} = \mu g\cos(9.3^\circ)\] | Dividing the maximum frictional force by the mass gives the maximum deceleration contribution from friction on the inclined plane. |
| 5 | \[a_{\text{gravity}} = g\sin(9.3^\circ)\] | When the car is moving uphill, the gravitational component along the slope, \(g\sin(9.3^\circ)\), also works to decelerate the car (acting downhill). |
| 6 | \[a_{\text{uphill}} = \mu g\cos(9.3^\circ) + g\sin(9.3^\circ)\] | The net deceleration is the sum of the deceleration from friction and the deceleration due to the gravitational component along the incline. |
| 7 | \[a_{\text{uphill}} = \left(\frac{3.80}{g}\right)g\cos(9.3^\circ) + g\sin(9.3^\circ) = 3.80\cos(9.3^\circ) + g\sin(9.3^\circ)\] | Substitute \(\mu = \frac{3.80}{g}\) from Step 1 into the net deceleration formula. |
| 8 | \[a_{\text{uphill}} = 3.80\cos(9.3^\circ) + 9.8\sin(9.3^\circ)\] | Using \(g = 9.8\, \text{m/s}^2\), the expression now contains numerical values and the trigonometric functions of \(9.3^\circ\). |
| 9 | \[a_{\text{uphill}} \approx 3.80(0.987) + 9.8(0.161) \approx 3.75 + 1.58 \approx 5.33 \, \text{m/s}^2\] | Evaluating \(\cos(9.3^\circ) \approx 0.987\) and \(\sin(9.3^\circ) \approx 0.161\) gives a net deceleration of about \(5.33\, \text{m/s}^2\). Since the deceleration is opposite to the direction of motion, it is expressed as a negative acceleration. |
| 10 | \[\boxed{a_{\text{uphill}} = -5.33 \, \text{m/s}^2}\] | This is the final result: when moving uphill on a \(9.3^\circ\) incline with the same static friction coefficient, the car decelerates at approximately \(-5.33 \, \text{m/s}^2\). |
Just ask: "Help me solve this problem."
Determine the force needed to push a \( 150 \) \( \text{kg} \) body up a smooth \( 30^\circ \) incline with an acceleration of \( 6 \) \( \text{m/s}^2 \).
A car slides up a frictionless inclined plane. How does the normal force of the incline on the car compare with the weight of the car?
A block of mass m is accelerated across a rough surface by a force of magnitude F exerted at an angle θ above the horizontal. The frictional force between the block and surface is ƒ. Find the acceleration of the block (as an equation).
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?
A horizontal \( 300 \) \( \text{N} \) force pushes a \( 40 \) \( \text{kg} \) object across a horizontal \( 10 \) \( \text{meter} \) frictionless surface. After this, the block slides up a \( 20^\circ \) incline. Assuming the incline has a coefficient of kinetic friction of \( 0.4 \), how far along the incline will the object slide?
\( 5.33 \, \text{m/s}^2 \). A negative number indicating deceleration is acceptable.
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 |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| 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] |
General Metric Conversion Chart
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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Try our free calculator to see what you need to get a 5 on the upcoming 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.
