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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\text{Coordinate System: } +x \text{ (East), } +y \text{ (North)}\] | Define the coordinate system with East as the positive \(x\) direction and North as the positive \(y\) direction. |
| 2 | \[F_{1x} = 170 \;\text{N}, \quad F_{1y} = 0 \;\text{N}\] | Student 1 pulls Eastward with \(170\,\text{N}\); hence, all force is in the \(x\) direction. |
| 3 | \[F_{2x} = 0 \;\text{N}, \quad F_{2y} = -100 \;\text{N}\] | Student 2 pulls Southward with \(100\,\text{N}\); therefore, the \(y\) component is negative. |
| 4 | \[F_{3x} = -200\sin(20^\circ), \quad F_{3y} = 200\cos(20^\circ)\] | Student 3 pulls with \(200\,\text{N}\) at \(20^\circ\) west of north. The \(y\) component is \(200\cos(20^\circ)\) (northward) and the \(x\) component is \(-200\sin(20^\circ)\) (westward). |
| 5 | \[F_{\text{net},x} = 170 – 200\sin(20^\circ)\] | Sum the \(x\) components: Student 1 contributes \(170\,\text{N}\) east, and Student 3 contributes \(-200\sin(20^\circ)\,\text{N}\) (west). |
| 6 | \[F_{\text{net},y} = -100 + 200\cos(20^\circ)\] | Sum the \(y\) components: Student 2 gives \(-100\,\text{N}\) (south) and Student 3 gives \(200\cos(20^\circ)\,\text{N}\) (north). |
| 7 | \(200\sin(20^\circ) \approx 68.4 \;\text{N}, \quad 200\cos(20^\circ) \approx 187.9 \;\text{N}\) | Calculate the approximate numerical values of the components for Student 3. |
| 8 | \[F_{\text{net},x} \approx 170 – 68.4 = 101.6 \;\text{N}\] | Compute the net \(x\) component using the approximated value. |
| 9 | \[F_{\text{net},y} \approx -100 + 187.9 = 87.9 \;\text{N}\] | Compute the net \(y\) component using the approximated value. |
| 10 | \[F_{\text{net}} = \sqrt{(101.6)^2 + (87.9)^2} \approx 134.4 \;\text{N}\] | Find the magnitude of the net force using the Pythagorean theorem. |
| 11 | \[\theta = \tan^{-1}\left(\frac{87.9}{101.6}\right) \approx 40.9^\circ\]\] | Determine the direction of the net force measured as the angle north of east. |
| 12 | \[\boxed{134.4 \;\text{N},\; 40.9^\circ \; \text{north of east}}\] | State the final net force magnitude and its direction. |
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Find the downward acceleration of an elevator, given that the ratio of a person’s stationary weight to their weight in the elevator is \(5:4\).
When a falling meteoroid is at a distance above the Earth’s surface of \( 3.00 \) times the Earth’s radius, what is its acceleration due to the Earth’s gravitation?
A child on Earth has a weight of \(500 \, \text{N}\). Determine the weight of the child if the Earth were to triple in both mass and radius (\(3M\) and \(3r\)).
A ball of mass \( m \) is suspended from two strings of unequal length as shown above. The magnitudes of the tensions \( T_1 \) and \( T_2 \) in the strings must satisfy which of the following relations?
A westward–moving car is changing its speed. The net force on the car ____.
A \(2,000 \, \text{kg}\) car collides with a stationary \(1,000 \, \text{kg}\) car. Afterwards, they slide \(6 \, \text{m}\) before coming to a stop. The coefficient of friction between the tires and the road is \(0.7\). Find the initial velocity of the \(2,000 \, \text{kg}\) car before the collision?
A block of weight \( W \) is pulled along a horizontal surface at constant speed by a force \( F \), which acts at an angle of \( \theta \) with the horizontal. The normal force exerted on the block by the surface has magnitude:
A person pulls a rope with a force \( F \) at an angle of \( 60^\circ \) to the horizontal. The rope is connected to a load over a frictionless pulley as shown in the diagram. The load is stationary. Which of the following is correct about the weight of the load and the net force exerted on the pulley by the rope?
According to Newton’s third law, each team in a tug of war pulls with equal force on the other team. What, then, determines which team will win?
A \( 240 \) \( \text{kg} \) block is dropped from \( 3.0 \) meters onto a spring, compresses the spring and comes to rest.
\(\boxed{134.4\,\text{N}\text{ at }40.9^\circ\text{ north of east}}\)
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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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