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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ mg – T = m a \] | For the hanging mass \(m\), the net force is its weight \(mg\) minus the tension \(T\), which equals \(m a\) by Newton’s second law. |
| 2 | \[ T = M a \] | For the cart of mass \(M\) on the horizontal track, the only horizontal force is the tension \(T\), so its acceleration satisfies \(T = M a\). |
| 3 | \[ mg – T = m \left( \frac{T}{M} \right) \] | Substitute \(a = \frac{T}{M}\) from Step 2 into the equation for the hanging mass. |
| 4 | \[ mg = T + \frac{m}{M}T = T \left( 1 + \frac{m}{M} \right) \] | Rearrange the equation to group the tension terms together. |
| 5 | \[ T = \frac{m g}{1 + \frac{m}{M}} \] | Solve for \(T\) by dividing both sides by \(\left(1 + \frac{m}{M}\right)\). |
| 6 | \[ T = \frac{m g M}{M + m} \] | Simplify the expression by multiplying numerator and denominator by \(M\). |
| 7 | \[ T = \frac{1 \times g \times 3}{3 + 1} = \frac{3g}{4} \] | Substitute \(m = 1\,\text{kg}\) and \(M = 3\,\text{kg}\) into the formula. |
| 8 | \[ \boxed{T = \frac{3g}{4} \approx 7.35\,\text{N}} \] | Using \(g \approx 9.8\,\text{m/s}^2\), the tension in the string is approximately \(7.35\,\text{N}\). |
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Three blocks are stacked on top of one another. The top block has a mass of \( 4.6 \, \text{kg} \), the middle one has a mass of \( 1.2 \, \text{kg} \), and the bottom one has a mass of \( 3.7 \, \text{kg} \).
Identify and calculate any normal forces between the objects.
A ball of mass \( m \) is suspended from two strings of unequal length as shown above. The magnitudes of the tensions \( T_1 \) and \( T_2 \) in the strings must satisfy which of the following relations?
A box rests on the (frictionless) bed of a truck. The truck driver starts the truck and accelerates forward. The box immediately starts to slide toward the rear of the truck bed.
A \( 4700 \, \text{kg} \) truck carrying a \( 900 \, \text{kg} \) crate is traveling at \( 25 \, \text{m/s} \) to the right along a straight, level highway, as shown above. The truck driver then applies the brakes, and as it slows down, the truck travels \( 55 \, \text{m} \) in the next \( 3.0 \, \text{s} \). The crate does not slide on the back of the truck.
For linear motion the term “inertia” refers to the same physical concept of
A runner pushes against the track to sprint forward. Which two action–reaction FORCE pairs are involved? Select two letters.
Why is more fuel required for a spacecraft to travel from the Earth to the Moon than to return from the Moon to the Earth?
The elliptical orbit of a comet is shown above. Positions \(1\) and \(2\) are, respectively, the farthest and nearest positions to the Sun, and at position \(1\) the distance from the comet to the Sun is \(10\) times that at position \(2\). What is the ratio \(\dfrac{F_1}{F_2}\), the force on the comet at position \(1\) to the force on the comet at position \(2\)?
Two spherical objects have equal masses and experience a gravitational force of \( 25 \) \( \text{N} \) towards one another. Their centers are \( 36 \) \( \text{cm} \) apart. Determine each of their masses.
A cannon fires a cannonball forward. The recoil of the cannon is backward. Why doesn’t the cannon move backward as fast as the cannonball moves forward?
\(\boxed{T = \frac{3g}{4} \approx 7.35\,\text{N}}\)
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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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