AP Physics

Unit 4 - Energy

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# Part (a) Free Body Diagram Explanation

Please visualize or draw out the FBD as described:

Gravity ([katex] \vec{mg} [/katex]): Acts directly downwards. With a mass of [katex] m = 90 + 12 = 102 [/katex] kg.
Normal Force ([katex] \vec{N} [/katex]): Acts perpendicular to the surface of the incline.
Frictional Force ([katex] \vec{f} [/katex]): Acts parallel to the incline against the direction of motion during ascent and in the opposite direction of velocity during descent.
Component of Gravitational Force down the incline ([katex] \vec{mg} \sin(\theta) [/katex]): Helps in descending and resists during ascending. Here [katex] \theta = 30^\circ [/katex].

# Part (b) Calculation of Work Done by Friction to Stop the Bicycle

Step Derivation/Formula Reasoning
1 [katex] F_f = \mu_k N [/katex] Frictional force, [katex] F_f [/katex], is the product of the coefficient of kinetic friction, [katex] \mu_k [/katex], and the normal force, [katex] N [/katex].
2 [katex] N = mg \cos(\theta) [/katex] Normal force is the component of the gravitational force perpendicular to the incline.
3 [katex] F_f = 0.7 \times 102 \times 9.8 \times \cos(30^\circ) [/katex] Calculate [katex] F_f [/katex] using [katex] \mu_k = 0.7 [/katex], [katex] m = 102 [/katex] kg, [katex] g = 9.8 \, \text{m/s}^2 [/katex], and [katex] \theta = 30^\circ [/katex].
4 [katex] W_f = -F_f d [/katex] Work done by friction, [katex] W_f [/katex], is the product of the frictional force and the distance, [katex] d [/katex], over which it acts, with a negative sign indicating work done against the motion.
5 [katex] W_f = -0.7 \times 102 \times 9.8 \times \cos(30^\circ) \times 9 [/katex] Substitute values to calculate the work done. The distance [katex] d = 9 [/katex] m.
6 [katex] W_f \approx -5454 \, \text{J} [/katex] Calculated work done by friction; it’s negative as it opposes the direction of motion.

# Part (c) Explanation – Difficulty of Traveling Up vs. Down the Incline

Step Derivation/Formula Reasoning
1 [katex] F_{\text{gravity, down}} = mg \sin(\theta) [/katex] Component of gravitational force along the incline that assists in descending and opposes during ascending.
2 [katex] F_{\text{friction, up}} = \mu_s N [/katex] Static friction opposing the upward motion, which is higher due to increased force requirements.
3 [katex] F_{\text{net, up}} = F_{\text{gravity, down}} + F_{\text{friction, up}} [/katex] Sum of forces opposing the ascent, both the gravitational pull back down the incline and the frictional force.
4 [katex] F_{\text{net, down}} = F_{\text{gravity, down}} – F_{\text{friction, down}} [/katex] Net force during descent is reduced because friction (now kinetic) is less than static friction and gravity assists in motion.
5 Comparison The cyclist finds it harder to travel up due to higher net opposing force (more friction and gravity opposing motion).

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  1. FBD with 3 forces.
  2. [katex] W_f \approx -5454 \, \text{J} [/katex]
  3. The cyclist finds it harder to travel up due to higher net opposing force (more friction and gravity opposing motion).

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]

  3. Perform the multiplication: [katex]\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}[/katex]

  4. Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

  1. 1. Some answers may vary by 1% due to rounding.
  2. Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
  3. Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
  4. Bookmark questions you can’t solve to revisit them later
  5. 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

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