0 attempts
0% avg
UBQ Credits
Part (a): Free‐Body Diagram of the Crate
Step | Derivation/Formula | Reasoning |
---|---|---|
a1 | \( Weight = m_{\text{crate}} g \) | This is the gravitational force acting downward on the crate. |
a2 | \( N \) | The normal force from the truck bed acts upward and balances the gravitational force. |
a3 | \( f \) | Because the truck is decelerating (slowing down to the right), friction acts to the left on the crate to ensure it decelerates with the truck. |
Part (b): Magnitude of the Crate’s Acceleration
Step | Derivation/Formula | Reasoning |
---|---|---|
b1 | \( v_{\text{avg}} = \frac{\Delta x}{t} = \frac{55}{3} \approx 18.33\,\text{m/s} \) | The average speed is calculated using the displacement \(\Delta x = 55\,\text{m}\) over the time \(t = 3.0\,\text{s}\). |
b2 | \( v_{\text{avg}} = \frac{v_i+v_x}{2} \) | This relation holds for uniform acceleration, where \(v_i = 25\,\text{m/s}\) is the initial speed and \(v_x\) is the final speed. |
b3 | \( v_x = 2v_{\text{avg}} – v_i \approx 2(18.33) – 25 \approx 11.66\,\text{m/s} \) | Solve for the final speed \(v_x\) using the average speed relation. |
b4 | \( a = \frac{v_x – v_i}{t} = \frac{11.66 – 25}{3.0} \approx -4.45\,\text{m/s}^2 \) | The acceleration is determined from the change in velocity over the time interval. The negative sign indicates deceleration; the magnitude is \(4.45\,\text{m/s}^2\). |
Part (c): Minimum Coefficient of Friction to Prevent Sliding
Step | Derivation/Formula | Reasoning |
---|---|---|
c1 | \( f_{\text{req}} = m_{\text{crate}}\,|a| = 900 \times 4.45 \approx 4005\,\text{N} \) | This is the frictional force required to give the crate the same deceleration as the truck. |
c2 | \( N = m_{\text{crate}} g = 900 \times 9.8 \approx 8820\,\text{N} \) | The normal force is the weight of the crate. |
c3 | \( \mu_{\text{min}} = \frac{f_{\text{req}}}{N} = \frac{4005}{8820} \approx 0.45 \) | The minimum coefficient of static friction is the ratio of the required frictional force to the normal force. Since the crate does not slide, this friction is static. |
Part (d): Spring Extension if the Truck Bed is Frictionless
Step | Derivation/Formula | Reasoning |
---|---|---|
d1 | \( a = \frac{15 – 25}{10} = -1\,\text{m/s}^2 \) | According to the given guidance, the truck (and crate) decelerates from \(25\,\text{m/s}\) to \(15\,\text{m/s}\) in \(10\,\text{s}\); the magnitude of acceleration is \(1\,\text{m/s}^2\). |
d2 | \( F_{\text{spring}} = m_{\text{crate}}\,|a| = 900 \times 1 = 900\,\text{N} \) | This is the net force that the spring must exert on the crate to produce the acceleration. |
d3 | \( x = \frac{F_{\text{spring}}}{k} = \frac{900}{9200} \approx 0.098\,\text{m} \) | The extension of the spring is found using Hooke’s law, where \(k = 9200\,\text{N/m}\) is the spring constant. |
Part (e): Spring Force when the Truck is Moving at Constant Speed
Step | Derivation/Formula | Reasoning |
---|---|---|
e1 | \( a = 0 \) | At a constant speed of \(25\,\text{m/s}\), there is no acceleration. |
e2 | \( F_{\text{spring}} = m_{\text{crate}}\,a = 900 \times 0 = 0\,\text{N} \) | With no net force acting on the crate, the spring neither stretches nor compresses. |
e3 | \( x = 0 \) | The spring extension at constant speed is \(0\,\text{m}\), which is less than the extension when the truck was accelerating (or decelerating) in part (d). |
Just ask: "Help me solve this problem."
Two blocks, A and B, are connected by a light string that passes over a frictionless pulley. Block A, of mass \( 10 \) \( \text{kg} \), rests on a rough plane that makes an angle of \( 45^{\circ} \) with the horizontal, while block B, of mass \( 17 \) \( \text{kg} \), hangs vertically. Starting from rest, what is the minimum coefficient of static friction between block A and the plane required to keep the system in static equilibrium?
When a golf ball is dropped to the pavement, it bounces back up.
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 \(2 \, \text{kg}\) ball is swung in a vertical circle. The length of the string the ball is attached to is \(0.7 \, \text{m}\). It takes \(0.4 \, \text{s}\) for the ball to travel one revolution (assume the ball travels at constant speed).
List at least 2 everyday forces that are not conservative, and explain why they aren’t.
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] |
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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
We crafted the ultimate A.P Physics 1 course that simplifies everything so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.
A quick explanation
Credits are used to grade your FRQs and GQs. Pro users get unlimited credits.
Submitting counts as 1 attempt.
Viewing answers or explanations count as a failed attempts.
Phy gives partial credit if needed
MCQs and GQs are are 1 point each. FRQs will state points for each part.
Phy customizes problem explanations based on what you struggle with. Just hit the explanation button to see.
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.
NEW! PHY AI accurately solves all questions
🔥 Get up to 30% off Elite Physics Tutoring
🧠 NEW! Learn Physics From Scratch Self Paced Course
🎯 Need exam style practice questions?