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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta x = \frac{1}{2} a t^2\] | Use constant–acceleration kinematics with initial velocity \(v_i = 0\). |
| 2 | \[a = \frac{2\Delta x}{t^2}\] | Algebraically isolate \(a\). |
| 3 | \[a = \frac{2(10)}{(4.5)^2} = 0.99\,\text{m/s}^2\] | Insert \(\Delta x = 10\,\text{m}\) and \(t = 4.5\,\text{s}\) to find \(a \approx 0.99\,\text{m/s}^2\). |
| 4 | \[F_{1x}=100\cos20^\circ,\;F_{2x}=-400\cos40^\circ,\;F_{3x}=500\cos10^\circ\] | Resolve each force into horizontal components; left is negative, right positive. |
| 5 | \[F_{1x}=93.97\,\text{N},\;F_{2x}=-306.6\,\text{N},\;F_{3x}=492.4\,\text{N}\] | Evaluate the trigonometric products. |
| 6 | \[F_{x,\text{applied}} = 279.8\,\text{N}\] | Sum \(F_{1x}+F_{2x}+F_{3x}\) to get the net applied horizontal force. |
| 7 | \[F_{1y}=+100\sin20^\circ,\;F_{2y}=-400\sin40^\circ,\;F_{3y}=-500\sin10^\circ\] | Resolve forces into vertical components; upward is positive. |
| 8 | \[F_{1y}=+34.2\,\text{N},\;F_{2y}=-257.1\,\text{N},\;F_{3y}=-86.8\,\text{N}\] | Compute the numeric values. |
| 9 | \[F_{y,\text{applied}} = -309.7\,\text{N}\] | Add the vertical components to find a downward net external load of \(309.7\,\text{N}\). |
| 10 | \[N = mg + 309.7\,\text{N}\] | Vertical equilibrium requires \(N + F_{1y} = mg + |F_{2y}| + |F_{3y}|\); simplify to this expression for the normal force. |
| 11 | \[f_k = \mu_k N = 0.2\,(mg + 309.7)\] | Kinetic friction opposes motion with magnitude \(\mu_k N\). |
| 12 | \[F_{\text{net}} = F_{x,\text{applied}} – f_k\] | Net horizontal force equals applied force minus friction (leftward). |
| 13 | \[ma = 279.8 – 0.2(mg + 309.7)\] | Apply Newton’s second law \(\sum F_x = ma\). |
| 14 | \[m(a + 1.96) = 217.8\] | Use \(g = 9.8\,\text{m/s}^2\) and simplify: \(0.2g = 1.96\,\text{N/kg}\). |
| 15 | \[m = \frac{217.8}{a + 1.96}\] | Isolate the unknown mass. |
| 16 | \[m = \frac{217.8}{0.99 + 1.96} = 73.9\,\text{kg}\] | Insert \(a \approx 0.99\,\text{m/s}^2\) to calculate \(m\). |
| 17 | \[\boxed{m \approx 7.4 \times 10^{1}\,\text{kg}}\] | Express the mass to two significant figures. |
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You pull a box with a constant force across a frictionless table using an attached rope held horizontally. If you now pull the rope with the same force at an angle to the horizontal (with the box remaining flat on the table). Does the acceleration of the box increase, decrease, or remain the same if the rope is pulled at an angle? Explain.
A projectile has the least speed at what point in its path?
A block rests on a flat plane inclined at an angle of \(30^\circ\) with respect to the horizontal. What is the minimum coefficient of friction necessary to keep the block from sliding?
A horizontal \( 300 \) \( \text{N} \) force pushes a \( 40 \) \( \text{kg} \) object across a horizontal \( 10 \) \( \text{meter} \) frictionless surface. After this, the block slides up a \( 20^\circ \) incline. Assuming the incline has a coefficient of kinetic friction of \( 0.4 \), how far along the incline will the object slide?
The International Space Station travels at \( 7660 \, \text{m/s} \). Find the average velocity of the space station if it takes \( 90 \, \text{minutes} \) to make one full orbit around Earth.
A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s. Then it slows down at a steady rate of 2.0 m/s2 for 4.0 s. What is the final speed of the car?
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 \( (1.2\,W) \)?
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 \( 1.5 \) \( \text{kg} \) block is pushed to the right with just enough force to get it to move. The block is pushed for five seconds with this constant force, then the force is released and the block slides to a stop. If the coefficient of kinetic friction is \( 0.300 \) and the coefficient of static friction is \( 0.400 \), calculate the amount of time that passes from when the force is applied to when the block stops.
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?
\(74\,\text{kg}\)
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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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