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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]T_{\text{max}} = mg[/katex] | The maximum tension [katex]T_{\text{max}}[/katex] that the wire can withstand is equal to the weight of the heaviest load it can support without breaking. Here, [katex]m = 70.0 \, \text{kg}[/katex] (mass of the person) and [katex]g = 9.8 \, \text{m/s}^2[/katex] (acceleration due to gravity). |
| 2 | [katex]T_{\text{max}} = 70.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2[/katex] | Calculate the maximum tension [katex]T_{\text{max}}[/katex] using the product of mass and acceleration due to gravity. |
| 3 | [katex]T_{\text{max}} = 686 \, \text{N}[/katex] | Performing the multiplication yields the maximum tension value. |
| 4 |
[katex]T = m_{\text{load}}(g + a)[/katex] |
Using netwons 2nd law, add all the forces acting on the load being lifted. In this case the Tension and weight act in opposite directions, thus [katex]T – m_{\text{load}}g = m_{\text{load}}a[/katex]. Re-arrange the equation for [katex]T[/katex] and factor out [katex]m_{\text{load}}[/katex] |
| 5 | [katex]686 \, \text{N} = 45.0 \, \text{kg} \cdot (9.8 \, \text{m/s}^2 + a)[/katex] | Since the wire can just barely support [katex]686 \, \text{N}[/katex], set the tension required to lift the load equal to this maximum. |
| 6 | [katex]9.8 \, \text{m/s}^2 + a = \frac{686 \, \text{N}}{45.0 \, \text{kg}}[/katex] | Divide both sides by the mass of the load to solve for the acceleration. |
| 7 | [katex]a = \frac{686 \, \text{N}}{45.0 \, \text{kg}} – 9.8 \, \text{m/s}^2[/katex] | Solve for [katex]a[/katex] by subtracting gravity’s acceleration from the result of the division. |
| 8 | [katex]a \approx 5.4 \, \text{m/s}^2[/katex] | Calculate the value of [katex]a[/katex]. This is the maximum vertical acceleration that can be achieved without breaking the wire. |
Just ask: "Help me solve this problem."
A person whose weight is 4.92 × 102 N is being pulled up vertically by a rope from the bottom of a cave that is 35.2 m deep. The maximum tension that the rope can withstand without breaking is 592 N. What is the shortest time, starting from rest, in which the person can be brought out of the cave?
A pulley system consists of two blocks of mass 5 kg and 10 kg, connected by a rope of negligible mass that passes over a pulley of radius 0.1 meters and mass 2 kg. The pulley is free to rotate about its axis. The system is released from rest, and the block of mass 10 kg starts to move downwards. Assuming that the coefficient of kinetic friction between the pulley and the rope is 0.2, and neglecting air resistance, determine
In the diagram shown a 20 N force is applied to a block B (7 kg). Block A has a mass of 3 kg. Assume frictionless conditions.
Three blocks of masses 5, 4, and 3 kg are placed side by side in that order. A 25 N force applied on the 5 kg block accelerates all three blocks together to the right. Find the acceleration of the blocks and the normal force the 4 kg block exerts on the 3 kg block.
Why do raindrops fall with constant speed during the later stages of their descent?
[katex]a \approx 5.4 \, \text{m/s}^2[/katex]
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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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