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| Derivation or Formula | Reasoning |
|---|---|
| \[\vec{F}_{\text{horizontal}} = 0\] | On the frictionless bed, the box experiences no horizontal contact force; gravity and the normal act vertically, so their horizontal components are zero. |
| \[\Sigma \vec{F}_x = 0 \Rightarrow m\vec{a}_{\text{box}} = 0\] | From Newton’s second law, zero net horizontal force implies the horizontal acceleration of the box is zero. |
| \[\vec{a}_{\text{box, ground}} = 0\] | Thus the box maintains its initial horizontal velocity, which is zero relative to the ground. |
| \[\vec{v}_{\text{box, ground}} = \text{constant}\] | The box remains essentially “in place” while the truck accelerates forward underneath it, making the box appear to slide toward the rear of the truck bed. |
| \[\text{Truck rear reaches box} \Rightarrow \text{box leaves bed}\] | When the truck’s rear edge reaches the stationary box, the box loses support and then falls straight downward under gravity. |
| Derivation or Formula | Reasoning |
|---|---|
| \[\vec{a}_t = \text{truck’s forward acceleration}\] | Chris is in a non-inertial frame accelerating forward with the truck. |
| \[\vec{F}_{\text{pseudo}} = -m\vec{a}_t\] | To apply Newton’s laws in the accelerating frame, Chris must introduce a backward inertial (pseudo) force of magnitude \(m a_t\) on every mass, including the box. |
| \[\Sigma \vec{F}’_x = \vec{F}_{\text{pseudo}}\] | With no real horizontal forces, the only horizontal force in Chris’s frame is the pseudo force. |
| \[m\vec{a}’ = -m\vec{a}_t\] | Applying Newton’s second law in the truck frame (primed quantities) gives the box a backward acceleration. |
| \[\vec{a}’_{\text{box, truck}} = -\vec{a}_t\] | Chris therefore observes the box accelerate toward the rear of the truck with magnitude equal to the truck’s forward acceleration. Once it reaches the edge, it appears to “fly off” and then fall. |
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The speed of a \(40 \, \text{N}\) hockey puck, sliding across a level ice surface, decreases at the rate of \(0.61 \, \text{m/s}^2\). The coefficient of kinetic friction between the puck and ice is
A block of mass \(m\) is accelerated across a rough surface by a force of magnitude \(F\) exerted at an angle \(\theta\) above the horizontal. The frictional force between the block and surface is \(f\). Find the acceleration of the block (as an equation).
A westward–moving car is changing its speed. The net force on the car ____.
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 \( F = 12 \) \( \text{N} \):
A block of mass \( m \), acted on by a force \( F \) directed horizontally, slides up an inclined plane that makes an angle \( \theta \) with the horizontal. The coefficient of sliding friction between the block and the plane is \( \mu \).
A 1100 kg car accelerates from 32 m/s to 8.0 m/s in 4.0 sec. What amount of force was needed to slow it down?
When a golf ball is dropped to the pavement, it bounces back up.
A vehicle is moving at a speed of 12.3 m/s on a decline when the brakes of all four wheels are fully applied, causing them to lock. The slope of the decline forms an angle of 18.0 degrees with the horizontal plane. Given that the coefficient of kinetic friction between the tires and the road surface is 0.650.
A spring launches a \(4 \, \text{kg}\) block across a frictionless horizontal surface. The block then ascends a \(30^\circ\) incline with a kinetic friction coefficient of \(\mu_k = 0.25\), stopping after \(55 \, \text{m}\) on the incline. If the spring constant is \(800 \, \text{N/m}\), find the initial compression of the spring. Disregard friction while in contact with the spring.
If an object is moving, is it possible for the net force acting on it to be zero? Explain.
Mary: The box remains essentially “in place” while the truck accelerates forward underneath it, making the box appear to slide toward the rear of the truck bed. When the truck’s rear edge reaches the stationary box, the box loses support and then falls straight downward under gravity.
Chris: Observes the box accelerate toward the rear of the truck with magnitude equal to the truck’s forward acceleration. Once it reaches the edge, it appears to “fly off” and then fall.
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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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