| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (a) Accelerarion of the particle when its displacement is 6 m | ||
| 1 | \[F = ma\] | Newton’s second law relates force \( F \), mass \( m \), and acceleration \( a \). |
| 2 | \[a = \frac{F}{m}\] | Rearrange the formula to solve for acceleration. |
| 3 | \[a = \frac{4\, \text{N}}{0.20\, \text{kg}}\] | Substitute the force from the graph (4 N) and the mass (0.20 kg). |
| 4 | \[a = 20\, \text{m/s}^2\] | Calculate the acceleration. |
| (b) Time taken for the object to be displaced the first 12 m | ||
| 1 | \[\Delta x = v_i t + \frac{1}{2} a t^2\] | Using the kinematic equation with initial velocity \( v_i = 0 \). |
| 2 | \[12 = \frac{1}{2} \cdot 20 \cdot t^2\] | Substitute \( \Delta x = 12 \) m and \( a = 20 \text{ m/s}^2 \). |
| 3 | \[12 = 10 t^2\] | Simplify the equation. |
| 4 | \[t^2 = 1.2\] | Divide both sides by 10. |
| 5 | \[t = \sqrt{1.2}\] | Solve for \( t \). |
| 6 | \[t \approx 1.095\, \text{s}\] | Calculate the time taken. |
| (c) The amount of work done by the net force in displacing the object the first 12 m | ||
| 1 | \[W = F \Delta x\] | Work done \( W \) is the product of force and displacement. |
| 2 | \[W = 4 \times 12\] | Substitute \( F = 4 \text{ N} \) and \( \Delta x = 12 \text{ m} \). |
| 3 | \[W = 48 \text{ J}\] | Calculate the work done. |
| (d) The speed of the object at displacement \( x = 12 \text{ m} \) | ||
| 1 | \[v_x^2 = v_i^2 + 2a \Delta x\] | Use the kinematic equation with initial velocity \( v_i = 0 \). |
| 2 | \[v_x^2 = 0 + 2 \cdot 20 \cdot 12\] | Substitute \( a = 20 \text{ m/s}^2 \) and \( \Delta x = 12 \text{ m} \). |
| 3 | \[v_x^2 = 480\] | Calculate \( v_x^2 \). |
| 4 | \[v_x = \sqrt{480}\] | Solve for \( v_x \). |
| 5 | \[v_x \approx 21.9 \, \text{m/s}\] | Calculate the velocity. |
| (e) The final speed of the object at displacement \( x = 20 \text{ m} \) | ||
| 1 | \[W_{total} = W_{1} + W_{2}\] | Calculate total work done by summing areas under the \( F \) vs. \( x \) graph. |
| 2 | \[W_{1} = F_{1} \times \Delta x_{1} = 4 \times 12 = 48 \, \text{J}\] | The work done on the first section (rectangle 0 to 12 m). |
| 3 | \[W_{2} = \frac{1}{2} \cdot 4 \cdot 8 = 16 \, \text{J}\] | The work done on the second section (triangular area from 12 m to 20 m). |
| 4 | \[W_{total} = 48 + 16 = 64 \, \text{J}\] | Total work done. |
| 5 | \[\text{K.E.} = \frac{1}{2}m v_x^2\] | Relate total work done to kinetic energy gain. |
| 6 | \[64 = \frac{1}{2} \cdot 0.20 \cdot v_x^2\] | Substitute \( m = 0.20 \, \text{kg} \). |
| 7 | \[v_x^2 = 640\] | Solve for \( v_x^2 \). |
| 8 | \[v_x = \sqrt{640}\] | Solve for \( v_x \). |
| 9 | \[\boxed{v_x \approx 25.3 \, \text{m/s}}\] | Calculate the final speed at \( x = 20 \text{ m} \). |
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A constant force of strength \( 20 \) \( \text{N} \) acts on an object of mass \( 3 \) \( \text{kg} \) as it moves a distance of \( 4 \) \( \text{m} \). If this force is applied perpendicular to the \( 4 \) \( \text{m} \) displacement, the work done by the force is equal to:
An object has a mass of \(10 \, \text{kg}\). For each case below answer the questions and provide an example.
A small sphere hangs from a string attached to the ceiling of a uniformly accelerating train car. It is observed that the string makes an angle of \(37^\circ\) with respect to the vertical. The magnitude of the acceleration \(a\) of the train car is most nearly:
A car travels to right at constant velocity. The net force on the car is
A hypothetical planet has a radius \( 2.0 \) times that of Earth, but has the same mass. What is the acceleration due to gravity near its surface?
A mass moving with a constant speed \( u \) encounters a rough surface and comes to a stop. The mass takes a time \( t \) to stop after encountering the rough surface. The coefficient of dynamic friction between the rough surface and the mass is \( 0.40 \). Which of the following expressions gives the initial speed \( u \)?
A block of mass \(m\) is accelerated across a rough surface by a force of magnitude \(F\) exerted at an angle \(\theta\) above the horizontal. The frictional force between the block and surface is \(f\). Find the acceleration of the block (as an equation).
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)?
A \( 25.0 \) \( \text{kg} \) block is placed at the top of an inclined plane set at an angle of \( 35 \) degrees to the horizontal. The block slides down the \( 1.5 \) \( \text{m} \) slope at a constant rate. How much work did friction do on the block?
What is the mass of a dog that weighs \(58 \, \text{N}\) on Earth?
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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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