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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[T_{l} + T_{r} = 125 + 500\] | For vertical equilibrium, the sum of the upward chain tensions must equal the total downward weights (board plus person). |
| 2 | \[T_{l} + 250 = 625\] | Substitute the given value for the right chain tension (\(T_{r}=250\,\text{N}\)) and the weights. |
| 3 | \[T_{l} = 625 – 250 = 375\,\text{N}\] | Solve for the left chain tension by simplifying the equation. |
| 4 | \[\boxed{T_{l} = 375\,\text{N}}\] | The final answer for part (a) is the left chain tension. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[500\,x + 125(2) = 250(4)\] | Taking moments about the left end (\(x=0\)) eliminates the left tension. The person’s weight (\(500\,\text{N}\)) acts at an unknown distance \(x\), the board’s weight (\(125\,\text{N}\)) acts at its midpoint (2 m), and the right chain’s tension (\(250\,\text{N}\)) acts at 4 m. |
| 2 | \[500\,x + 250 = 1000\] | Simplify by calculating the moment due to the board’s weight (\(125 \times 2 = 250\)) and the moment due to the right chain (\(250 \times 4 = 1000\)). |
| 3 | \[500\,x = 1000 – 250 = 750\] | Isolate the term with \(x\) by subtracting 250 from 1000. |
| 4 | \[x = \frac{750}{500} = 1.5\,\text{m}\] | Solve for \(x\) to find the position of the person from the left end. |
| 5 | \[\boxed{x = 1.5\,\text{m}}\] | The final answer for part (b) is that the person is hanging 1.5 m from the left end of the board. |
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A box is sliding down an incline at a constant speed of \( 2 \, \text{m/s} \). The angle of the incline is \( \theta \). The magnitude of the total of the opposing forces is \( 16 \, \text{N} \). Derive an equation for the force of gravity acting on the box.
A \(1 \, \text{kg}\) mass and an unknown mass \(M\) hang on opposite sides of a pulley suspended from the ceiling. When the masses are released, \(M\) accelerates downward at \(5 \, \text{m/s}^2\). Find the value of \(M\).
A loop-de-loop roller coaster has a radius of \( 30 \) \( \text{m} \). Determine the apparent weight a \( 500 \) \( \text{N} \) person will feel at the bottom of the loop while traveling at a speed of \( 25 \) \( \text{m/s} \) and at the top of the loop while traveling at a speed of \( 20 \) \( \text{m/s} \).
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.
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 car is driving at \(25 \, \text{m/s}\) when a light turns red \(100 \, \text{m}\) ahead. The driver takes an unknown amount of time to react and hit the brakes, but manages to skid to a stop at the red light. If \(\mu_s = 0.9\) and \(\mu_k = 0.65\), what was the reaction time of the driver?
A person whose weight is \(4.92 \times 10^2 \, \text{N}\) is being pulled up vertically by a rope from the bottom of a cave that is \(35.2 \, \text{m}\) deep. The maximum tension that the rope can withstand without breaking is \(592 \, \text{N}\). What is the shortest time, starting from rest, in which the person can be brought out of the cave?
A \(3300 \, \text{m}\)-high mountain is located on the equator. How much faster does a climber on top of the mountain move than a surfer at a nearby beach? The Earth’s radius is \(6400 \, \text{km}\) and the Earth’s mass is \(5.97 \times 10^{24} \, \text{kg}\).
There are two cables that lift an elevator, each with a force of \(10{,}000 \, \text{N}\). The \(1{,}000 \, \text{kg}\) elevator is lifted from the first floor and accelerates over \(10 \, \text{m}\) until it reaches its top speed of \(6 \, \text{m/s}\). What is the mass of the people in the elevator?
A small block slides without friction along a track toward a circular loop. The block has more than enough speed to remain firmly in contact with the track as it goes around the loop. The magnitude of the block’s acceleration at the top of the loop is
a) \( 375 \, \text{N} \)
b) \( 1.5 \, \text{m} \) from the left end of the board
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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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