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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[10g – T_{2} = 10a\] | For the 10 kg block (which moves downward), the net force is its weight minus the rope tension \(T_{2}\) and equals \(10a\). |
| 2 | \[T_{1} – 5g = 5a\] | For the 5 kg block (which moves upward), the net force is the rope tension \(T_{1}\) minus its weight and equals \(5a\). |
| 3 | \[T_{2}=T_{1}\,e^{\mu\pi}\] | Since the rope slips on the pulley, the capstan relation applies over a contact angle of \(\pi\) radians with \(\mu=0.2\); thus, the tension on the heavy side is \(T_{2}=T_{1}\,e^{0.2\pi}\). |
| 4 | \[T_{1}=5g+5a\] | Solve the 5 kg block equation for \(T_{1}\). |
| 5 | \[10g -e^{0.2\pi}(5g+5a)=10a\] | Substitute \(T_{2}=e^{0.2\pi}(5g+5a)\) into the 10 kg block equation. |
| 6 | \[10g-5e^{0.2\pi}g = a(10+5e^{0.2\pi})\] | Distribute and gather like terms in \(a\) and the constants. |
| 7 | \[a = \frac{5g(2-e^{0.2\pi})}{5(2+e^{0.2\pi})} = \frac{g(2-e^{0.2\pi})}{2+e^{0.2\pi}}\] | Simplify the expression to obtain \(a\) in terms of \(g\) and \(e^{0.2\pi}\). |
| 8 | \[a \approx \frac{9.8(2-1.875)}{2+1.875} \approx \frac{9.8(0.125)}{3.875} \approx 0.32\,\text{m/s}^2\] | Using \(e^{0.2\pi}\approx1.875\) and \(g\approx9.8\,\text{m/s}^2\), we find the acceleration. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[T_{1}=5g+5a\] | From the 5 kg block equation, solve for \(T_{1}\); substitute \(a\approx0.32\,\text{m/s}^2\) and \(g=9.8\,\text{m/s}^2\). |
| 2 | \[T_{1}\approx5(9.8)+5(0.32)=49+1.6\approx50.6\,\text{N}\] | Calculate \(T_{1}\) numerically. |
| 3 | \[T_{2}=T_{1}\,e^{0.2\pi}\approx50.6\times1.875\approx94.7\,\text{N}\] | Apply the capstan relation to find \(T_{2}\) on the 10 kg side. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[f = T_{2}-T_{1}\] | The difference in tensions is provided by the kinetic friction at the pulley–rope interface. |
| 2 | \[f \approx 94.7-50.6 \approx 44.1\,\text{N}\] | Substitute the computed tensions to find the frictional force magnitude. |
| 3 | \[\text{Direction: Counterclockwise}\] | Since the 10 kg block moves downward (causing clockwise tendency), the kinetic friction acts opposite to the rope’s slip, i.e. it exerts a counterclockwise force on the pulley. |
Just ask: "Help me solve this problem."
A race car travels in a circular track of radius \( 200 \) \( \text{m} \). If the car moves with a constant speed of \( 80 \) \( \text{m/s} \),
In lacrosse, a typical throw is made by rotating the stick through an angle of roughly 90°, then releasing the ball when the stick is vertical, as shown above. If the 1 meter long stick is at rest when horizontal and the ball leaves the stick with a velocity of 10 m/s, what angular acceleration must the stick experience?
The angular velocity of an electric motor is \( \omega = \left(20 – \frac{1}{2} t^2 \right) \, \text{rad/s} \), where \(t\) is in seconds.
A ladder at rest is leaning against a wall at an angle. Which of the following forces must have the same magnitude as the frictional force exerted on the ladder by the floor?
A system consists of two small disks, of masses \( m \) and \( 2m \), attached to a rod of negligible mass of length \( 3l \) as shown above. The rod is free to turn about a vertical axis through point \( P \). The two disks rest on a rough horizontal surface; the coefficient of friction between the disks and the surface is \( \mu \). At time \( t = 0 \), the rod has an initial counterclockwise angular velocity \( \omega_0 \) about \( P \). The system is gradually brought to rest by friction. Develop expressions for the following quantities in terms of \( \mu \), \( m \), \( l \), \( g \), and \( \omega_0 \).
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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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