Step | Derivation/Formula | Reasoning |
---|---|---|
1 | U_{\text{spring}} = \frac{1}{2} k x^2 | The potential energy stored in the compressed spring, where k is the spring constant and x is the compression distance. |
2 | W_{\text{friction}} = \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d | Work done against friction while moving up the incline: \mu_k is the coefficient of kinetic friction, m is mass, g is acceleration due to gravity, \theta is the angle of incline, and d is the distance traveled up the incline. |
3 | PE = m \cdot g \cdot h | The gravitational potential energy gained by the block at the end of the incline. Here, h is the vertical height, calculated further in Step 5. |
4 | U_{\text{spring}} = W_{\text{friction}} + PE | Conservation of mechanical energy: The total mechanical energy (spring potential energy) initially equals the sum of the work done against friction and the gravitational potential energy at the end. |
5 | h = d \cdot \sin(\theta) | Using the distance d traveled up the incline and the angle \theta to calculate the height h . |
6 | Substituting h , W_{\text{friction}} , and PE in step 4: | \frac{1}{2} k x^2 = \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d + m \cdot g \cdot d \cdot \sin(\theta) |
7 | Simplifying and solving for x : | x^2 = \frac{2(\mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d + m \cdot g \cdot d \cdot \sin(\theta))}{k} x = \sqrt{\frac{2d(mg(\mu_k \cdot \cos(\theta) + \sin(\theta)))}{k}} |
8 | x = \sqrt{\frac{2 \cdot 55 \cdot (4 \cdot 9.8 \cdot (0.25 \cdot \cos(30^\circ) + \sin(30^\circ)))}{800}} | Substitute values for d = 55 \, \text{m} , m = 4 \, \text{kg} , g = 9.8 \, \text{m/s}^2 , \mu_k = 0.25 , \theta = 30^\circ , and k = 800 \, \text{N/m} to find x . |
9 | x = \boxed{1.965\, m} | Final value. |
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A bullet of mass 0.0500 kg traveling at 50.0 m/s is fired horizontally into a wooden block suspended from a long rope. The mass of the wooden block is 0.300 kg and it is initially at rest. The collision is completely inelastic and after impact the bullet+ wooden block move together until the center of mass of the system rises a vertical distance h above its initial position.
An object is projected vertically upward from ground level. It rises to a maximum height H . If air resistance is negligible, which of the following must be true for the object when it is at a height H/2 ?
A ball is thrown straight up. At what point does the ball have the most energy?
The speed of a 40 N hockey puck, sliding across a level ice surface, decreases at the rate of 0.61 m/s2. The coefficient of kinetic friction between the puck and ice is
A 84.4 kg climber is scaling the vertical wall. His safety rope is made of a material that behaves like a spring that has a spring constant of 1.34 x 103 N/m. He accidentally slips and falls 0.627 m before the rope runs out of slack. How much is the rope stretched when it breaks his fall and momentarily brings him to rest?
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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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