0 attempts
0% avg
UBQ Credits
Objective: Determine the acceleration of the blocks and the tension in the cord given the masses, coefficients of friction, and incline angle given:
Part a: Calculate the acceleration of the blocks
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] F_{\text{gravity, parallel A}} = m_A g \sin(\theta) [/katex] | Parallel component of gravitational force for A. |
| 2 | [katex] F_{\text{gravity, parallel B}} = m_B g \sin(\theta) [/katex] | Parallel component of gravitational force for B. |
| 3 | [katex] F_{\text{friction A}} = \mu_A m_A g \cos(\theta) [/katex] | Frictional force on A. |
| 4 | [katex] F_{\text{friction B}} = \mu_B m_B g \cos(\theta) [/katex] | Frictional force on B. |
| 5 | [katex] F_{\text{net A}} = F_{\text{gravity, parallel A}} – F_{\text{friction A}} [/katex] | Net force on A. |
| 6 | [katex] F_{\text{net B}} = F_{\text{gravity, parallel B}} – F_{\text{friction B}} [/katex] | Net force on B. |
| 7 | [katex] F_{\text{net}} = F_{\text{net B}} – F_{\text{net A}} [/katex] | Total net force on the system. |
| 8 | [katex] a = \frac{F_{\text{net}}}{m_A + m_B} [/katex] | Acceleration of the system. |
Plug in the given values:
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 9 | [katex] a = \frac{(m_B g \sin(\theta) – \mu_B m_B g \cos(\theta)) – (m_A g \sin(\theta) – \mu_A m_A g \cos(\theta))}{m_A + m_B} [/katex] | Substitute the net forces from steps 5 and 6. |
| 10 | [katex] a = \frac{(5 \cdot 9.8 \cdot \sin(32^\circ) – 0.30 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ)) – (5 \cdot 9.8 \cdot \sin(32^\circ) – 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ))}{5 + 5} [/katex] | Substitute given values. |
| 11 | [katex] a = \frac{(5 \cdot 9.8 \cdot (\sin(32^\circ) – 0.30 \cdot \cos(32^\circ))) – (5 \cdot 9.8 \cdot (\sin(32^\circ) – 0.20 \cdot \cos(32^\circ)))}{10} [/katex] | Simplify the expression. |
| 12 | [katex] a = \frac{5 \cdot 9.8 \cdot (0.10 \cdot \cos(32^\circ))}{10} [/katex] | Combine like terms. |
| 13 | [katex] a = \frac{9.8 \cdot (0.10 \cdot \cos(32^\circ))}{2} [/katex] | Simplify further. |
Part b: Calculate the tension in the cord
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] T = m_A \cdot a + F_{\text{friction A}} [/katex] | Tension equals the force to accelerate block A plus frictional force on A. |
| 2 | [katex] F_{\text{friction A}} = \mu_A \cdot m_A \cdot g \cdot \cos(\theta) [/katex] | Frictional force opposing the motion of block A. |
| 3 | [katex] T = m_A \cdot a + \mu_A \cdot m_A \cdot g \cdot \cos(\theta) [/katex] | Substitute the frictional force into the tension formula. |
| 4 | [katex] T = m_A \cdot \left( \frac{F_{\text{net}}}{m_A + m_B} \right) + \mu_A \cdot m_A \cdot g \cdot \cos(\theta) [/katex] | Substitute the expression for � from the acceleration calculation. |
| 5 | [katex] T = 5 \cdot \left( \frac{(5 \cdot 9.8 \cdot \sin(32^\circ) – 0.30 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ)) – (5 \cdot 9.8 \cdot \sin(32^\circ) – 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ))}{10} \right) + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) [/katex] | Insert given values for masses, coefficients of friction, gravitational acceleration, and angle. |
| 6 | [katex] T = 5 \cdot \left( \frac{5 \cdot 9.8 \cdot (0.10 \cdot \cos(32^\circ))}{10} \right) + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) [/katex] | Simplify the expression for the net force component. |
| 7 | [katex] T = \frac{5 \cdot 9.8 \cdot (0.10 \cdot \cos(32^\circ))}{2} + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) [/katex] | Further simplify the tension formula. |
| 8 | [katex] T = \frac{5 \cdot 9.8 \cdot 0.10 \cdot \cos(32^\circ)}{2} + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) [/katex] | Combine like terms for the final tension calculation. |
| 9 | [katex] T \approx 17.66 , \text{N} [/katex] | Calculate the numeric value for tension. |
[katex] \boxed{T = 17.66 , \text{N}} [/katex]
Just ask: "Help me solve this problem."
A 25.0-kg box is released on a 23.5° incline and accelerates down the incline at 0.35 m/s2. Find the friction force impeding its motion. What is the coefficient of kinetic friction?
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
A student is watching their hockey puck slide up and down an incline. They give the puck a quick push along a frictionless table, and it slides up a 30° rough incline (µk = .4) of distance d, with an initial speed of 5 m/s, and then it slides back down.
A ring is pulled on by three forces. If the ring is not moving, how big is the force [katex]F[/katex]?
A \(30 \, \text{g}\) bullet is fired with a speed of \(500 \, \text{m/s}\) into a wall.
By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
