Step | Formula Derivation | Reasoning |
---|---|---|
1 | PE = KE + W_{\text{friction}} | Conservation of Energy |
2 | -\frac{1}{2}mv^2 + mgh = F_{\text{friction}}d | Plug in formulas for each type of energy and solve for Fd (work). |
3 | F_{\text{friction}} = \mu mg\cos(\theta) | Friction force is the product of the coefficient of friction, normal force, and gravity. |
4 | -\frac{1}{2}mv^2 + mgh = \mu mg\cos(\theta)d | Substitute friction force from Step 3. |
5 | sin(\theta) = h/d | Solve for h in terms of d and plug into the equation above |
5 | v = \sqrt{2g(dsin\theta) + 2\mu g\cos(\theta)s} | Rearrange and solve for v. |
The snowboarder’s speed at the base of the incline using energy considerations is \boxed{~25.5 , \text{m/s}}.
Now, let’s use this speed to calculate the distance traveled on the level surface.
Step | Formula Derivation | Reasoning |
---|---|---|
1 | W_{\text{friction}} = F_{\text{friction}} \cdot d | Work done by friction is the product of friction force and distance (d). |
2 | KE = W_{\text{friction}} | Conservation of Energy |
3 | \frac{1}{2}mv^2 = \mu mgd | Substitute KE expression W. |
4 | d = \frac{\frac{1}{2}mv^2}{\mu mg} | Solve for d. |
The distance the snowboarder travels along the level surface, calculated using energy considerations is \boxed{267.60 , \text{m}}.
Phy can also check your working. Just snap a picture!
The figure shows a graph of the position x of two cars, C and D, as a function of time t. According to this graph, which statements about these cars must be true? (There could be more
than one correct choice.)
Which of the following graphs represent an object at rest?
Ball 1 is dropped from rest at time t = 0 from a tower of height h. At the same instant, ball 2 is launched upward from the ground with the initial speed v_0. If air resistance is negligible, at what time t will the two balls pass each other?
An object of mass 2 kg is thrown vertically downwards with an initial kinetic energy of 100 J. What is the distance fallen by the object at the instant when its kinetic energy has doubled?
An object is thrown downward at 23 m/s from the top of a 200 m tall building.
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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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