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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( T_{\text{bottom}} = m_{\text{bottom}} \cdot g \) | The tension in the bottom rope is equal to the weight of the bottom block, since the only force it needs to balance is the gravitational force acting on the \$3.0 \, \text{kg}\$ block. |
| 2 | \( T_{\text{bottom}} = 3.0 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 \) | Substitute the mass of the bottom block and the acceleration due to gravity into the equation. |
| 3 | \( T_{\text{bottom}} = 29.4 \, \text{N} \) | The tension in the bottom rope is calculated to be \(29.4 \, \text{N}\). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( T_{\text{top}} = (m_{\text{top}} + m_{\text{bottom}}) \cdot g \) | The tension in the top rope must balance the total weight of both blocks. |
| 2 | \( 63.0 = (m_{\text{top}} + 3.0) \cdot 9.8 \) | Substitute the known values: \( T_{\text{top}} = 63.0 \, \text{N} \), \( m_{\text{bottom}} = 3.0 \, \text{kg} \), and \( g = 9.8 \, \text{m/s}^2 \). |
| 3 | \( \frac{63.0}{9.8} = m_{\text{top}} + 3.0 \) | Solve for \( m_{\text{top}} \) by dividing both sides by the acceleration due to gravity. |
| 4 | \( m_{\text{top}} = \frac{63.0}{9.8} – 3.0 \) | Isolate \( m_{\text{top}} \) on one side of the equation. |
| 5 | \( m_{\text{top}} = 3.4286 \, \text{kg} \approx 3.43 \, \text{kg} \) | The mass of the top block is calculated to be approximately \(3.43 \, \text{kg}\). |
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From the top of a \( 74.0 \) \( \text{m} \) high building, a \( 1.00 \) \( \text{kg} \) ball is dropped in the presence of air resistance. The ball reaches the ground with a speed of \( 31.0 \) \( \text{m/s} \), indicating that drag was significant. How much energy was lost in the form of air resistance/drag during the fall?
The heaviest train ever pulled by a single engine was over [katex] 2 \, \text{km} [/katex] long. A force of [katex] 1.13 \times 10^5 \, \text{N} [/katex] is needed to get the train to start moving. If the coefficient of static friction is [katex] 0.741 [/katex] and the coefficient of kinetic friction is [katex] .592 [/katex], what is the train’s mass?
Suppose an object is accelerated by a force of \( 100 \) \( \text{N} \). Suddenly a second force of \( 100 \) \( \text{N} \) in the opposite direction is exerted on the object, so that the forces cancel. The object
A hockey puck glides on perfectly frictionless ice at constant velocity. Which statement is true?
A car is going through a dip in the road whose curvature approximates a circle of radius \( 200 \) \( \text{m} \). At what velocity will the occupants of the car appear to weigh \( 20\% \) more than their normal weight \( (1.2\,W) \)?
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
Two objects (49.0 and 24.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley hangs from the ceiling. Find the acceleration of the objects and the tension in the string.
A person stands on a scale in an elevator. His apparent weight will be the greatest when the elevator
When the brakes of an automobile are applied, the road exerts the greatest retarding force
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 \( 20 \) \( \text{m/s} \).
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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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