AP Physics

Unit 4 - Energy

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# Part (a): Time Elapsed from Leaving the Table to Hitting the Floor

Step Derivation/Formula Reasoning
1 [katex] t = \sqrt{\frac{2h}{g}} [/katex] Since the motion in the y-direction is a free fall, we use the kinematic equation [katex] y = \frac{1}{2}gt^2 [/katex] for the vertical motion. Solving for [katex] t [/katex] gives the time it takes to fall a distance [katex] h [/katex].

# Part (b): Horizontal Component of the Velocity of the Block Just Before It Hits the Floor

Step Derivation/Formula Reasoning
1 [katex] v_x = \frac{D}{t} [/katex] Using the result from part (a) and substituting [katex] t = \sqrt{\frac{2h}{g}} [/katex], we get [katex] v_x = \frac{D}{\sqrt{\frac{2h}{g}}} = \sqrt{\frac{D^2g}{2h}} [/katex]. This is the velocity necessary to cover horizontal distance [katex] D [/katex] in time [katex] t [/katex].
2 [katex] v_x = \frac{D}{\sqrt{\frac{2h}{g}}} [/katex] Replace [katex] t [/katex] with the equation found in part a.

# Part (c): Work Done on the Block by the Spring

Step Derivation/Formula Reasoning
1 [katex] El = KE [/katex] The work done by the spring (elastic energy) is transforms into kinetic energy. Since we found the velocity in the previous part we can solve for the kinetic energy.
2  [katex] El = KE  = \frac{1}{2}mv^2[/katex] Formula for kinetic energy
3 [katex] KE = \frac{1}{2} m \left(\frac{D}{\sqrt{\frac{2h}{g}}}\right)^2 [/katex] Substitute in velocity found from previous step.
4 [katex] KE  = \frac{mgD^2}{4h}[/katex] Simplify equation. Note that the Kinetic energy is the work done by the spring.

# Part (d): Spring Constant

Step Derivation/Formula Reasoning
1 [katex]  \frac{1}{2}kx^2 = \frac{1}{2}mv^2 [/katex] The spring energy ([katex]EL[/katex]) is equal to the kinetic energy as mentioned in part c. Hence we can set [katex] EL = KE [/katex] and solve for [katex]k[/katex].
2 [katex]  k = \frac{mv^2}{x^2}[/katex] Solve for k
3 [katex] k = \frac{m \left(\frac{D}{\sqrt{\frac{2h}{g}}}\right)^2}{x^2} [/katex] Substitute in the [katex] v [/katex], to get the final equation in terms of [katex] M, x, D, h, [/katex]
4 [katex] k = \frac{mD^2 g}{2hx^2} [/katex] Simplify

These steps address each part of the query based on principles of mechanics, conservation of energy, and kinematic equations.

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  1. [katex] t = \sqrt{\frac{2h}{g}} [/katex]
  2. [katex] v_x = \frac{D}{\sqrt{\frac{2h}{g}}} [/katex]
  3. [katex] KE  = \frac{mgD^2}{4h}[/katex]
  4. [katex] k = \frac{mD^2 g}{2hx^2} [/katex]

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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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