Step | Derivation/Formula | Reasoning |
---|---|---|
1 | PE_{\text{initial}} = 0 | The object is initially on the ground, so its initial potential energy is zero (since potential energy is relative to height above a reference point). |
2 | PE_{\text{max}} = mgH | At maximum height H , all the kinetic energy has been converted to potential energy. Potential energy at height H is given by mgH , where m is the mass, g is the acceleration due to gravity, and H is the height. |
3 | PE_{\text{half}} = mg \frac{H}{2} | At half the maximum height \frac{H}{2} , the potential energy is mg \frac{H}{2} . |
4 | KE_{\text{initial}} = mgH | Using the principle of conservation of mechanical energy, the initial kinetic energy must equal the potential energy at maximum height, since there is no kinetic energy at that point (the object is momentarily stationary). |
5 | KE_{\text{half}} = KE_{\text{initial}} – PE_{\text{half}} | To find the kinetic energy at height \frac{H}{2} , subtract the potential energy at that height from the total initial mechanical energy. |
6 | KE_{\text{half}} = mgH – mg \frac{H}{2} = mg \frac{H}{2} | Simplifying, the kinetic energy at \frac{H}{2} is mg \frac{H}{2} , or half the initial kinetic energy. |
7 | ME_{\text{initial}} = ME_{\text{half}} = KE_{\text{initial}} | Total mechanical energy is conserved (since air resistance is negligible). The mechanical energy at any point during the motion is the sum of potential and kinetic energies and is equal to the initial total mechanical energy. |
8 | (a) False, (b) False, (c) False, (d) True, (e) False | Evaluating each statement: (a) Potential energy at \frac{H}{2} is not half of the initial (it’s \frac{1}{2} mgH ). (b) Speed is not directly proportional in this manner. (c) Total mechanical energy is constant, not halved. (d) Kinetic energy at \frac{H}{2} is indeed half of the initial kinetic energy, as shown in the calculations. (e) Since (d) is true, (e) is false. |
Phy can also check your working. Just snap a picture!
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An object undergoes constant acceleration. Starting from rest, the object travels 5 meters in the first second. Then it travels 15 meters in the next second. What total distance will be covered after the 3rd second?
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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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