| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( F = ma \) | Using Newton’s Second Law, the force applied on an object is equal to the mass of the object times its acceleration. |
| 2 | \( 4.5 \, \text{N} = m \times 2.5 \, \text{m/s}^2 \) | Substitute the given force and acceleration to find the mass of the initial block. |
| 3 | \( m = \frac{4.5 \, \text{N}}{2.5 \, \text{m/s}^2} \) | Rearranging the equation to solve for mass \( m \) of the first block. |
| 4 | \( m = 1.8 \, \text{kg} \) | Calculate to find the mass of the first block. The mass is \( 1.8 \, \text{kg} \). |
| 5 | \( m_{\text{total}} = 1.8 \, \text{kg} + 4.0 \, \text{kg} \) | Sum the two masses since the second block is placed on top of the first block without sliding. |
| 6 | \( m_{\text{total}} = 5.8 \, \text{kg} \) | Calculate the total mass of the system with both blocks. |
| 7 | \( a_{\text{new}} = \frac{F}{m_{\text{total}}} \) | Using Newton’s Second Law again, solve for the new acceleration \( a_{\text{new}} \) with the total mass. |
| 8 | \( a_{\text{new}} = \frac{4.5 \, \text{N}}{5.8 \, \text{kg}} \) | Substitute the force and total mass to find the new acceleration. |
| 9 | \( a_{\text{new}} = 0.78 \, \text{m/s}^2 \) | Calculate the new acceleration of the system, boxed as the final answer. |
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Traveling at a speed of 15.9 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.659. What is the speed of the automobile after 1.59 s have elapsed? Ignore the effects of air resistance.
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?
The cart with mass \( M = 3 \, \text{kg} \) is pulled by a massless string and moving on a horizontal track. A weight with mass \( m = 1 \, \text{kg} \) is hung from the other end of the string through a pulley system. Due to the gravitational force acting on the weight of mass \( m \), the cart is accelerated to the left. Find the tension in the string.
A truck is traveling at \(35 \, \text{m/s}\) when the driver realizes the truck has no brakes. He sees a ramp off the road, inclined at \(20^\circ\), and decides to go up it to help the truck come to a stop. How far does the truck travel before coming to a stop (assume no friction)?
An elevator carrying a person of mass \( m \) is moving upward and slowing down. How does the magnitude \( F \) of the force exerted on the person by the elevator floor compare with the magnitude \( mg \) of the gravitational force?
A 135.0 N force is applied to a 30.0 kg box at 42 degree angle to the horizontal. If the force of friction is 85.0, what is the net force and acceleration? If the object starts from rest, how far has it traveled in 3.3 sec?
A uniform rope of weight \( 30 \, \text{N} \) hangs from a hook. A box of mass \( 40 \, \text{kg} \) is suspended from the rope. What is the tension in the rope?
A net force of \( 8.0 \) \( \text{N} \) accelerates a \( 4.0 \) \( \text{kg} \) body from rest to a speed of \( 5.0 \) \( \text{m s}^{-1} \). Which of the following is equal to the work done by the force?
A person whose weight is \(4.92 \times 10^2 \, \text{N}\) is being pulled up vertically by a rope from the bottom of a cave that is \(35.2 \, \text{m}\) deep. The maximum tension that the rope can withstand without breaking is \(592 \, \text{N}\). What is the shortest time, starting from rest, in which the person can be brought out of the cave?
An object weighs \( 432 \) \( \text{N} \) on the surface of Earth. At a height of \( 3R_{\text{Earth}} \) above Earth’s surface, what is its weight?
\( .78 \, \text{m/s}^2 \)
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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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