Step | Formula Derivation | Reasoning |
---|---|---|
1 | k = \frac{mg}{x} | Calculate spring constant k using initial conditions. m = 3.0 , \text{kg} , g = 9.8 , \text{m/s}^2 , x = 0.12 , \text{m} . |
2 | k = \frac{(3.0 , \text{kg})(9.8 , \text{m/s}^2)}{0.12 , \text{m}} = 245 , \text{N/m} | Substitute values to find k . |
3 | x_{\text{new}} = \frac{m_{\text{new}}g}{k} | Calculate the new equilibrium position x_{\text{new}} for the 4.0 kg block. m_{\text{new}} = 4.0 , \text{kg} . |
4 | x_{\text{new}} = \frac{(4.0 , \text{kg})(9.8 , \text{m/s}^2)}{245 , \text{N/m}} = 0.16 , \text{m} | Substitute values to find x_{\text{new}} . |
5 | \Delta x_{\text{total}} = 2x_{\text{new}} | Total distance fallen is twice the new equilibrium position. x_{\text{new}} is known. |
6 | \Delta x_{\text{total}} = 2 \times 0.16 , \text{m} \times 100 , \frac{\text{cm}}{\text{m}} = 32 , \text{cm} | Convert the total distance to centimeters. \Delta x_{\text{total}} in meters to cm. |
The total distance the 4.0 kg block falls before its direction is reversed is 32 cm.
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A horizontal force of 110 N is applied to a 12 kg object, moving it 6 m on a horizontal surface where the kinetic friction coefficient is 0.25. The object then slides up a 17° inclined plane. Assuming the 110 N force is no longer acting on the incline, and the coefficient of kinetic friction there is 0.45, calculate the distance the object will slide on the incline.
Two boxes are tied together by a string and are sitting at rest on a frictionless surface. Between the two boxes is a massless compressed spring. The string trying the two boxes is then cut and the spring expands, pushing the boxes apart. The box on the left has four times the mass of the box on the right.
From the figure above, determine the which characteristic fits this collision best.
You are working out on a rowing machine. Each time you pull the rowing bar toward you, it moves a distance of 1.25 m in a time of 0.84 s. The readout on the display indicates that the average power you are producing is 76 W. What is the magnitude of the force that you exert on the handle?
A 81 kg student dives off a 45 m tall bridge with an 18 m long bungee cord tied to his feet and to the bridge. You can consider the bungee cord to be a flexible spring. What spring constant must the bungee cord have for the student’s lowest point to be 2.0 m above the water?
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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_1m_2}{r^2} |
a = \frac{\Delta v}{\Delta t} | f = \mu N |
R = \frac{v_i^2 \sin(2\theta)}{g} |
Circular Motion | Energy |
---|---|
F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
a_c = \frac{v^2}{r} | PE = mgh |
KE_i + PE_i = KE_f + PE_f |
Momentum | Torque and Rotations |
---|---|
p = m v | \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 |
---|
F = -k x |
T = 2\pi \sqrt{\frac{l}{g}} |
T = 2\pi \sqrt{\frac{m}{k}} |
Constant | Description |
---|---|
g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
\mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
k | Spring constant, in \text{N/m} |
M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
Variable | SI Unit |
---|---|
s (Displacement) | \text{meters (m)} |
v (Velocity) | \text{meters per second (m/s)} |
a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
t (Time) | \text{seconds (s)} |
m (Mass) | \text{kilograms (kg)} |
Variable | Derived SI Unit |
---|---|
F (Force) | \text{newtons (N)} |
E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
P (Power) | \text{watts (W)} |
p (Momentum) | \text{kilogram meters per second (kgm/s)} |
\omega (Angular Velocity) | \text{radians per second (rad/s)} |
\tau (Torque) | \text{newton meters (Nm)} |
I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |
f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \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}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|
Pico- | p | 10^{-12} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
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