| Derivation or Formula | Reasoning |
|---|---|
| \[ W = F\,\Delta x\,\cos(\theta) \] | This is the general formula for work done by a constant force, where \(F\) is the force magnitude, \(\Delta x\) is the displacement, and \(\theta\) is the angle between the force and displacement. |
| \[ \theta = 90^\circ \] | Because the force is applied perpendicular to the displacement, the angle \(\theta\) between the force and displacement is \(90^\circ\). |
| \[ \cos(90^\circ) = 0 \] | The cosine of \(90^\circ\) is zero, which means that the entire product for work will be zero. |
| \[ W = 20\,\text{N} \times 4\,\text{m} \times 0 = 0\,\text{J} \] | Substituting the given values into the work formula shows that the work done is \(0\,\text{J}\) since the cosine factor nullifies the contribution of the force. |
| \[ \boxed{0\,\text{J}} \] | This is the final answer. The force, being perpendicular to the displacement, does no work, so options (b) through (e) are incorrect because they ignore the \(\cos(90^\circ)=0\) factor. |
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Two blocks, [katex] m_2 > m_1 [/katex], having the same kinetic energy, move from a frictionless surface onto a surface having friction coefficient [katex] \mu_k [/katex]. Which block will travel further before stopping.
A linear spring of force constant \( k \) is used in a physics lab experiment. A block of mass \( m \) is attached to the spring and the resulting frequency, \( f \), of the simple harmonic oscillations is measured. Blocks of various masses are used in different trials, and in each case, the corresponding frequency is measured and recorded. If \( f^{2} \) is plotted versus \( \frac{1}{m} \), the graph will be a straight line with slope
List at least 2 everyday forces that are not conservative, and explain why they aren’t.
A projectile of mass 0.750 kg is shot straight up with an initial speed of 18.0 m/s.
A \( 7.3 \) \( \text{kg} \) mass is placed on a spring with a spring constant of \( 34 \) \( \text{N/cm} \). How much does this stretch the spring?
You kick a ball straight up. Compare the sign of the work done by gravity on the ball while it goes up with the sign of the work done by gravity while it goes down.
A block of mass [katex] m [/katex] is moving on a horizontal frictionless surface with a speed [katex] v_0 [/katex] as it approaches a block of mass [katex] 2m [/katex] which is at rest and has an ideal spring attached to one side.
When the two blocks collide, the spring is completely compressed and the two blocks momentarily move at the same speed, and then separate again, each continuing to move.
In the figure above, the marble rolls down the track and around a loop-the-loop of radius \( R \). The marble has mass \( m \) and radius \( r \). What minimum height \( h_{min} \) must the track have for the marble to make it around the loop-the-loop without falling off? Express your answer in terms of the variables \( R \) and \( r \).
Two balls are dropped from the roof of a building. One ball has twice as massive as the other and air resistance is negligible. Just before hitting the ground, the more massive ball has ball ____ the kinetic energy of the less massive ball.
A pendulum bob of mass m on a cord of length L is pulled sideways until the cord makes an angle [katex] \theta [/katex] with the vertical. The change in potential energy of the bob during the displacement is:
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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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