| 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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The heaviest train ever pulled by a single engine was over \( 2 \, \text{km} \) long. A force of \( 1.13 \times 10^5 \, \text{N} \) is needed to get the train to start moving. If the coefficient of static friction is \( 0.741 \) and the coefficient of kinetic friction is \( .592 \), what is the train’s mass?
A \( 1.0 \, \text{kg} \) lump of clay is sliding to the right on a frictionless surface with a speed of \( 2 \, \text{m/s} \). It collides head-on and sticks to a \( 0.5 \, \text{kg} \) metal sphere that is sliding to the left with a speed of \( 4 \, \text{m/s} \). What is the kinetic energy of the combined objects after the collision?
An object is moving at constant velocity. Which of the following could be the free-body diagram representing the forces acting on the object?
A vehicle is moving at a speed of 12.3 m/s on a decline when the brakes of all four wheels are fully applied, causing them to lock. The slope of the decline forms an angle of 18.0 degrees with the horizontal plane. Given that the coefficient of kinetic friction between the tires and the road surface is 0.650.
How does the time t1 of a block m reaching the bottom of slide 1 compare with t2, the time taken block of mass 2m to reach the end of slide 2 that’s curved? The blocks are released from the same height.
When the brakes of an automobile are applied, the road exerts the greatest retarding force
An object of mass \( m = 3.0 \) \( \text{kg} \) is attached to one end of a string with negligible mass and length \( L = 0.80 \) \( \text{m} \). The object is released from rest at time \( t = 0 \), when the string is horizontal. At time \( t = t_1 \) the object is at the location shown in the figure, where the string is vertical. Which of the following is most nearly the magnitude of the tension in the string at time \( t = t_1 \)?
The gravitational force that the Moon exerts on Earth is often cited as the source of the tides we witness. However, the gravitational force the Sun exerts on Earth is over \(100\) times greater than the force the Moon exerts on Earth.
Why is the force from the Moon credited for the tides, and not the force from the Sun?
Three blocks of masses \( 1.0 \, \text{kg} \), \( 2.0 \, \text{kg} \), and \( 4.0 \, \text{kg} \) are connected by massless strings, one of which passes over a frictionless pulley of negligible mass, as shown above. Calculate each of the following.
The graph below is a plot of position versus time. For which labeled segments is the velocity positive and the acceleration negative?
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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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