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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(W_{\text{box}} = mg\) | Calculate the weight of the box. The weight is determined by multiplying the mass of the box by the acceleration due to gravity \(g = 10 \, \text{m/s}^2\). |
| 2 | \(W_{\text{box}} = 40 \, \text{kg} \times 10 \, \text{m/s}^2 = 400 \, \text{N}\) | Calculate the numerical value of the weight, \(W_{\text{box}}\). The weight of the box is 400 N. |
| 3 | \(T_{\text{bottom}} = W_{\text{box}} = 400 \, \text{N}\) | The tension at the bottom of the rope is equal to the weight of the box, 400 N, as it supports the box. |
| 4 | \(W_{\text{rope}} = 30 \, \text{N}\) | Given that the rope itself has a weight of 30 N. |
| 5 | \(T_{\text{top}} = W_{\text{rope}} + W_{\text{box}}\) | The tension at the top of the rope must support both the weight of the box and the weight of the rope. |
| 6 | \(T_{\text{top}} = 30 \, \text{N} + 400 \, \text{N} = 430 \, \text{N}\) | Calculate the tension at the top of the rope by adding the weight of the rope to the weight of the box. |
| 7 | It varies from 400 \, \text{N} at the bottom of the rope to 430 \, \text{N} at the top | The tension varies along the length of the rope, being greatest at the top where the entire weight of both the rope and box are supported. |
– (a) is incorrect because it does not account for the additional force from the box.
– (b) is incorrect because it treats tension as solely the weight of the box.
– (c) is incorrect as it oversimplifies the tensions without considering both rope and box weights.
– (d) is incorrect because it does not account for varying tensions.
– (e) is the correct answer; it accounts for the combined weight of the rope and the box and shows variation.
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Imagine a hypothetical planet that has two moons. Moon \(\#1\) is in a circular orbit of radius \(R\) and has a mass \(M\).
Three students are pulling on a bag of skittles. Each is pulling with a horizontal force. If student 1 pulls Eastward with [katex]170 \, \text{N}[/katex], student 2 pulls Southward with [katex]100 \, \text{N}[/katex] and student 3 pulls with [katex]200 \, \text{N}[/katex] at an angle of [katex]20^\circ [/katex] west of north, what is the net force caused by the three students on the bag of skittles?
A westward–moving car is changing its speed. The net force on the car ____.
A student is watching their hockey puck slide up and down an incline. They give the puck a quick push along a frictionless table, and it slides up a \( 30^\circ \) rough incline (\( \mu_k = 0.4 \)) of distance \( d \), with an initial speed of \( 5 \) \( \text{m/s} \), and then it slides back down.
A 45 kg crate accelerates at 1.65 m/s2 when pulled by a rope with a force of 200 N. Find the angle the rope is pulled at. Friction is negligible.
A \( 15 \) \( \text{N} \) force is pushing a \( 40 \) \( \text{N} \) block down a incline. The angle of the inline is \( \alpha = 40^{\circ} \). The coefficient of static friction between the block and the incline is \( \mu_s = 0.75 \) and the coefficient of kinetic friction is \( \mu_k = 0.65 \).
A block is given a brief push so that it slides up a ramp. After the block reaches its highest point, it slides back down, but the magnitude of its acceleration is less on the descent than on the ascent. Why?
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?
What is the mass of a dog that weighs \(58 \, \text{N}\) on Earth?
A person stands on a scale in an elevator. His apparent weight will be the greatest when the elevator
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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] |
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] |
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