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."
Why is the stopping distance of a truck much shorter than for a train going the same speed? Hint: try deriving a formula or stopping distance.
A 135.0 N force is applied to a 30.0 kg box at 42 degree angle to the horizontal. If the force of friction is 85.0, what is the net force and acceleration? If the object starts from rest, how far has it traveled in 3.3 sec?
A “doomsday” asteroid with a mass of \( 1010 \, \text{kg} \) is hurtling through space. Unless the asteroid’s speed is changed by about \( 0.20 \, \text{cm/s} \), it will collide with Earth and cause tremendous damage. Researchers suggest that a small “space tug” sent to the asteroid’s surface could exert a gentle constant force of \( 2.5 \, \text{N} \). For how long must this force act?
A \( 1 \) \( \text{kg} \) mass on a \( 37^{\circ} \) incline is connected to a \( 3.0 \) \( \text{kg} \) mass on a horizontal surface, as shown. The surfaces and the pulley are frictionless. If \( F = 12 \) \( \text{N} \):
An object has a mass of 10 kg. For each case below answer the questions and provide an example.
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] | |
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
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.
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.
