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Part a: Calculate the acceleration of the system
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] F_{\text{net,A}} = m_A \cdot a [/katex] | Net force on mass A equals mass times acceleration. |
| 2 | [katex] F_{\text{net,B}} = m_B \cdot a [/katex] | Net force on mass B equals mass times acceleration. |
| 3 | [katex] F_{\text{net,A}} = T – m_A \cdot g [/katex] | Tension upwards minus weight of A downwards. |
| 4 | [katex] F_{\text{net,B}} = m_B \cdot g – T [/katex] | Weight of B downwards minus tension upwards. |
| 5 | [katex] m_A \cdot a = T – m_A \cdot g [/katex] | Substitute step 1 into step 3. |
| 6 | [katex] m_B \cdot a = m_B \cdot g – T [/katex] | Substitute step 2 into step 4. |
| 7 | [katex] m_A \cdot a + m_B \cdot a = m_B \cdot g – m_A \cdot g [/katex] | Add step 5 and step 6 equations. |
| 8 | [katex] a = \frac{(m_B – m_A) \cdot g}{m_A + m_B} [/katex] | Solve for acceleration . |
Use the given number from the problem.
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 9 | [katex] a = \frac{(2.4 – 3.2) \cdot 9.8}{3.2 + 2.4} [/katex] | Plug in known values. |
| 10 | [katex] a = \frac{-0.8 \cdot 9.8}{5.6} [/katex] | Simplify the numerator and denominator. |
| 11 | [katex] a = -1.4 , \text{m/s}^2 [/katex] | Calculate the acceleration. |
[katex] \boxed{a = -1.4 , \text{m/s}^2} [/katex] Negative sign indicates downward direction.
Part b: Calculate the tension in the string
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] T = m_A \cdot (g + a) [/katex] | Tension equals mass A times (gravity plus acceleration). |
Using the previously calculated acceleration:
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 2 | [katex] T = 3.2 \cdot (9.8 – 1.4) [/katex] | Plug in known values for mass and acceleration |
| 3 | [katex] T = 3.2 \cdot 8.4 [/katex] | Simplify the terms in the parentheses. |
| 4 | [katex] T = 26.88 , \text{N} [/katex] | Calculate the tension. |
[katex] \boxed{T = 26.88 , \text{N}} [/katex]
Part c: Calculate the final speed of mass A before it hits the ground
| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | [katex] v = \sqrt{2 \cdot} [/katex] | a |
| 2 | [katex] v = \sqrt{2 \cdot 1.4 \cdot 0.5} [/katex] | Plug in the magnitude of and . |
| 3 | [katex] v = \sqrt{1.4} [/katex] | Calculate the expression under the square root. |
| 4 | [katex] v = 1.18 , \text{m/s} [/katex] | Find the square root to get the final velocity. |
[katex] \boxed{v = 1.18 , \text{m/s}} [/katex]
Just ask: "Help me solve this problem."
A 1000 kg car is traveling east at 20m/s when it collides perfectly inelastically with a northbound 2000 kg car traveling at 15m/s. If the coefficient of kinetic friction is 0.9, how far, and at what angle do the two cars skid before coming to a stop?
Two blocks made of different materials, connected by a thin cord, slide down a plane ramp inclined at an angle [katex] \theta [/katex] to the horizontal. If the coefficients of friction are µA = .2 and µB = .3 and if mA = mB = 5.0 kg determine:
A sled moves with constant speed down a sloped hill. The angle of the hill with respect to the horizontal is 10.0°. What is the coefficient of kinetic friction between the sled and the hill’s surface?
The alarm at a fire station rings and a 79.34-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 4.20 m). Just before landing, his speed is 1.36 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?
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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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