| Step | Formula Derivation | Reasoning |
|---|---|---|
| 1 | F_{\text{parallel}} = mg\sin(\theta) | Force parallel to the incline, where m is mass, g is acceleration due to gravity, and \theta is the incline angle. |
| 2 | F_{\text{normal}} = mg\cos(\theta) | Normal force, perpendicular to the incline. |
| 3 | F_{\text{friction}} = \mu_k F_{\text{normal}} | Frictional force, where \mu_k is the coefficient of kinetic friction. |
| 4 | F_{\text{friction}} = 0.250 \times mg\cos(25^\circ) | Substitute values for \mu_k and F_{\text{normal}}. |
| 5 | F_{\text{pull}} = F_{\text{parallel}} + F_{\text{friction}} | Total force to pull the crate includes parallel force and frictional force. |
| 6 | F_{\text{pull}} = mg\sin(25^\circ) + 0.250 \times mg\cos(25^\circ) | Combine parallel and frictional forces. |
| 7 | W_{\text{output}} = F_{\text{parallel}} \times d | Work output, where F_{\text{parallel}} is the force parallel to the incline and d is the distance. |
| 8 | W_{\text{output}} = mg\sin(25^\circ) \times 2.5 | Work done against gravity. |
| 9 | W_{\text{input}} = F_{\text{pull}} \times d | Work input, force to pull the crate. |
| 10 | W_{\text{input}} = \left( mg\sin(25^\circ) + 0.250 \times mg\cos(25^\circ) \right) \times 2.5 | Work done to pull the crate including overcoming friction. |
| 11 | \text{Efficiency} = \frac{W_{\text{output}}}{W_{\text{input}}} \times 100% | Efficiency formula. |
| 12 | \text{Efficiency} = \frac{mg\sin(25^\circ) \times 2.5}{\left( mg\sin(25^\circ) + 0.250 \times mg\cos(25^\circ) \right) \times 2.5} \times 100% | Substitute W_{\text{output}} and W_{\text{input}}. |
| 13 | \text{Efficiency} = \frac{\sin(25^\circ)}{\sin(25^\circ) + 0.250 \times \cos(25^\circ)} \times 100% | Simplify by canceling mg and d. |
| 14 | \text{Efficiency} = 65.10\% | Calculated efficiency. |
Phy can also check your working. Just snap a picture!
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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.
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 1.0-kg object is moving with a velocity of 6.0 m/s to the right. It collides and sticks to a 2.0-kg object moving with a velocity of 3.0 m/s in the same direction. How much kinetic energy was lost in the collision?
A small block moving with a constant speed v collides inelastically with a block M attached to one end of a spring k. The other end of the spring is connected to a stationary wall. Ignore friction between the blocks and the surface.
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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} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
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