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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.
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| 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] |
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