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To solve part (a), calculate the work performed by the kinetic frictional force acting on the skis.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]KE_i = \frac{1}{2}mv_i^2[/katex] | Calculate the initial kinetic energy (KE) using the mass [katex] m = 58 \, \text{kg} [/katex] and the initial velocity [katex] v_i = 7.2 \, \text{m/s} [/katex]. |
| 2 | [katex]KE_i = \frac{1}{2} \times 58 \times (7.2)^2 = 1503.36 \, \text{J}[/katex] | Substitute the values into the kinetic energy formula. |
| 3 | [katex]KE_f = \frac{1}{2}mv_f^2[/katex] | Calculate the final kinetic energy using the final velocity [katex] v_f = 3.8 \, \text{m/s} [/katex]. |
| 4 | [katex]KE_f = \frac{1}{2} \times 58 \times (3.8)^2 = 418.76 \, \text{J}[/katex] | Substitute the values into the kinetic energy formula. |
| 5 | [katex]W_{\text{gravity}} = mgh [/katex] | Calculate the work done by gravity, where [katex] h = d \sin(\theta) [/katex] is the height gained climbing the incline. [katex] d = 2.3 \, \text{m} [/katex] and [katex] \theta = 28^\circ [/katex]. |
| 6 | [katex]h = 2.3 \sin(28^\circ) = 1.08 \, \text{m}[/katex] | Calculate the vertical height climbed using [katex] \sin(28^\circ) \approx 0.4695 [/katex]. |
| 7 | [katex]W_{\text{gravity}} = 58 \times 9.8 \times 1.080 = 614 \, \text{J}[/katex] | Substitute [katex] g = 9.8 \, \text{m/s}^2 [/katex]. |
| 8 | [katex] KE_i = KE_f + W_f + PE [/katex] | Place all energy transformation into a single conservation of energy equation: The initial kinetic energy transforms into the final kinetic energy, work done by friction, and the potential energy of the skier. |
| 9 | [katex]W_f = 1503.36 \, – \, 418.76 \, – \, 614 [/katex] | Plug in all values and solve for work done by friction [katex] W_f [/katex]. |
| 10 | [katex] W_f = 470.6 \, \text{J}[/katex] | The negative sign indicates that the work done by friction is in the direction opposite to the motion. |
To solve part (b), determine the magnitude of the kinetic frictional force.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]W = f_k \times d \times \cos(\theta)[/katex] | Work done by a force [katex] f_k [/katex] over a distance [katex] d [/katex], where [katex] \theta [/katex] is the angle between the force and the displacement (which in the case of friction, is [katex]180^\circ[/katex]). |
| 2 | [katex] 470.6 = -f_k \times 2.3 \times \cos(180^\circ)[/katex] | Substitute the work calculated from part (a) and the distance [katex]2.3 \, \text{m}[/katex]. [katex] \cos(180^\circ) = -1 [/katex]. |
| 3 | [katex] -470.6 = f_k \times 2.3[/katex] | Simplify the equation. |
| 4 | [katex]f_k = \frac{-470.6}{2.3}[/katex] | Solve for [katex]f_k[/katex]. |
| 5 | [katex]f_k = -204.6 \, \text{N}[/katex] | The magnitude of the kinetic frictional force. |
Just ask: "Help me solve this problem."
A kickball is rolled by the pitcher at a speed of 10 m/s and it is kicked by another student. The kickball deforms a little during the kick, and then rebounds with a velocity of 15 m/s as its shape restores to a perfect sphere. Select all that must be true about the kickball and the kicking foot system.
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.
One end of a spring is attached to a solid wall while the other end just reaches to the edge of a horizontal, frictionless tabletop, which is a distance [katex] h [/katex] above the floor. A block of mass M is placed against the end of the spring and pushed toward the wall until the spring has been compressed a distance [katex] x [/katex]. The block is released and strikes the floor a horizontal distance [katex] D [/katex] from the edge of the table. Air resistance is negligible.
Derive an expressions for the following quantities only in terms of [katex] M, x, D, h, [/katex] and any constants.
A car is driving at \(25 \, \text{m/s}\) when a light turns red \(100 \, \text{m}\) ahead. The driver takes an unknown amount of time to react and hit the brakes, but manages to skid to a stop at the red light. If \(\mu_s = 0.9\) and \(\mu_k = 0.65\), what was the reaction time of the driver?
Two objects (49.0 and 24.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley hangs from the ceiling. Find the acceleration of the objects and the tension in the string.
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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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