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Part (a): Free‐Body Diagram of the Crate
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| a1 | \( Weight = m_{\text{crate}} g \) | This is the gravitational force acting downward on the crate. |
| a2 | \( N \) | The normal force from the truck bed acts upward and balances the gravitational force. |
| a3 | \( f \) | Because the truck is decelerating (slowing down to the right), friction acts to the left on the crate to ensure it decelerates with the truck. |
Part (b): Magnitude of the Crate’s Acceleration
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| b1 | \( v_{\text{avg}} = \frac{\Delta x}{t} = \frac{55}{3} \approx 18.33\,\text{m/s} \) | The average speed is calculated using the displacement \(\Delta x = 55\,\text{m}\) over the time \(t = 3.0\,\text{s}\). |
| b2 | \( v_{\text{avg}} = \frac{v_i+v_x}{2} \) | This relation holds for uniform acceleration, where \(v_i = 25\,\text{m/s}\) is the initial speed and \(v_x\) is the final speed. |
| b3 | \( v_x = 2v_{\text{avg}} – v_i \approx 2(18.33) – 25 \approx 11.66\,\text{m/s} \) | Solve for the final speed \(v_x\) using the average speed relation. |
| b4 | \( a = \frac{v_x – v_i}{t} = \frac{11.66 – 25}{3.0} \approx -4.45\,\text{m/s}^2 \) | The acceleration is determined from the change in velocity over the time interval. The negative sign indicates deceleration; the magnitude is \(4.45\,\text{m/s}^2\). |
Part (c): Minimum Coefficient of Friction to Prevent Sliding
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| c1 | \( f_{\text{req}} = m_{\text{crate}}\,|a| = 900 \times 4.45 \approx 4005\,\text{N} \) | This is the frictional force required to give the crate the same deceleration as the truck. |
| c2 | \( N = m_{\text{crate}} g = 900 \times 9.8 \approx 8820\,\text{N} \) | The normal force is the weight of the crate. |
| c3 | \( \mu_{\text{min}} = \frac{f_{\text{req}}}{N} = \frac{4005}{8820} \approx 0.45 \) | The minimum coefficient of static friction is the ratio of the required frictional force to the normal force. Since the crate does not slide, this friction is static. |
Part (d): Spring Extension if the Truck Bed is Frictionless
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| d1 | \( a = \frac{15 – 25}{10} = -1\,\text{m/s}^2 \) | According to the given guidance, the truck (and crate) decelerates from \(25\,\text{m/s}\) to \(15\,\text{m/s}\) in \(10\,\text{s}\); the magnitude of acceleration is \(1\,\text{m/s}^2\). |
| d2 | \( F_{\text{spring}} = m_{\text{crate}}\,|a| = 900 \times 1 = 900\,\text{N} \) | This is the net force that the spring must exert on the crate to produce the acceleration. |
| d3 | \( x = \frac{F_{\text{spring}}}{k} = \frac{900}{9200} \approx 0.098\,\text{m} \) | The extension of the spring is found using Hooke’s law, where \(k = 9200\,\text{N/m}\) is the spring constant. |
Part (e): Spring Force when the Truck is Moving at Constant Speed
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| e1 | \( a = 0 \) | At a constant speed of \(25\,\text{m/s}\), there is no acceleration. |
| e2 | \( F_{\text{spring}} = m_{\text{crate}}\,a = 900 \times 0 = 0\,\text{N} \) | With no net force acting on the crate, the spring neither stretches nor compresses. |
| e3 | \( x = 0 \) | The spring extension at constant speed is \(0\,\text{m}\), which is less than the extension when the truck was accelerating (or decelerating) in part (d). |
Just ask: "Help me solve this problem."
When a basketball is dropped to the pavement, it bounces back up. Is a force needed to make it bounce back up? If so, what exerts the force?
Block m1 is stacked on top of block m2. Block m2 is connected by a light cord to block m3, which is pulled along a frictionless surface with a force F as shown in the diagram above. Block m1 is accelerated at the same rate as block m2 because of the frictional forces between the two blocks. If all three blocks have the same mass m, what is the minimum coefficient of static friction between block m1 and block m2?
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
The moment of inertia of a uniform solid sphere (mass \( M \), radius \( R \)) about a diameter is \( \frac{2}{5}MR^2 \). The sphere is placed on an inclined plane (angle \( \theta \)) and released from rest.
A 100 kg person is riding a 10 kg bicycle up a 25° hill. The hill is long and the coefficient of static friction is 0.9. The person rides 10 m up the hill then takes a rest at the top. If she then starts from rest from the top of the hill and rolls down a distance of 7 m before squeezing hard on the brakes locking the wheels. How much work is done by friction to bring the bicycle to a full stop, knowing that the coefficient of kinetic friction is 0.65?
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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] |
General Metric Conversion Chart
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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