Step | Formula Derivation | Reasoning |
---|---|---|
1 | E_{\text{total}} = E_{\text{kinetic}} + E_{\text{potential}} | Total mechanical energy in a spring-mass system is conserved, comprising kinetic and potential energy. |
2 | E_{\text{potential}} = \frac{1}{2}kx^2 | Potential energy in a spring, where k is the spring constant and x is the displacement from equilibrium. |
3 | E_{\text{kinetic}} = \frac{1}{2}mv^2 | Kinetic energy of the mass, where m is the mass and v is the velocity. |
4 | E_{\text{total, initial}} = \frac{1}{2}kx^2 | Total energy initially (at maximum stretch) is all potential energy. Given: k = 25 , \text{N/m}, x = 52.0 , \text{cm} = 0.52 , \text{m}. |
5 | E_{\text{total, halfway}} = E_{\text{kinetic, halfway}} + E_{\text{potential, halfway}} | Total energy halfway to equilibrium. |
6 | E_{\text{potential, halfway}} = \frac{1}{2}k\left(\frac{x}{2}\right)^2 | Potential energy halfway to equilibrium (x/2). |
7 | E_{\text{total, initial}} = E_{\text{total, halfway}} | Conservation of mechanical energy. |
8 | \frac{1}{2}kx^2 = \frac{1}{2}mv^2 + \frac{1}{2}k\left(\frac{x}{2}\right)^2 | Equate initial and halfway total energies. |
9 | v = \sqrt{\frac{kx^2 – k\left(\frac{x}{2}\right)^2}{m}} | Solve for v. Given: m = 80 , \text{g} = 0.080 , \text{kg}. |
Let’s calculate the velocity of the mass when it is halfway to the equilibrium position.
Step | Formula Derivation | Reasoning |
---|---|---|
10 | v = 7.96 , \text{m/s} | Velocity of the mass halfway to the equilibrium position. |
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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.
A proton (mp = 1.67 x10-27 kg) is being accelerated along a straight line at 3.6 ×1015 m/s2 in a machine. The proton has an initial speed of 2.4 x107 m/s and travels 3.5 cm.
A mechanic pushes a 2500 \, \text{kg} car from rest to a final speed v by doing 5.0 \times 10^3 \, \text{J} of work on the car. Frictional effect between the car and the ground are negligible. What is the final speed of the car?
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.
How does the speed v1 of a block m reaching the bottom of slide 1 compare with v2, the speed of a block 2m reaching the end of slide 2? The blocks are released from the same height.
7.96 m/s
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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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