AP Physics Unit

Unit 4 - Energy

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Mathematical

MCQ

A crate is pulled 2.5 m at constant velocity along a 25° incline. The coefficient of kinetic friction between the crate and the plane is 0.250. What is the efficiency of this procedure?

  1. 35%

  2. 45%

  3. 55%

  4. 65%

  5. 75%

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.

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Nerd-Notes.com
KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_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 MotionEnergy
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
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients 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.
kSpring 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
VariableSI 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)}
VariableDerived 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. 

  1. Start with the given measurement: \text{5 km}

  2. 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}}

  3. 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}

  4. 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}

  1. Some answers may be slightly off by 1% depending on rounding, etc.
  2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
  3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
  4. Bookmark questions that you can’t solve so you can come back to them later. 
  5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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