AP Physics Unit

Unit 4 - Energy

Advanced

Mathematical

FRQ

A snowboarder starts from rest and slides down a 32° incline that’s 75 m long.

  1. (a) If the coefficient of friction is 0.12, what is the snowboarder’s speed at the base of the incline? (3 points)
  2. (b) At the foot of the incline the snow levels off and has the same coefficient of friction. Using only energy find how far the snowboarder travels along the level surface. (4 points)
Step Formula Derivation Reasoning
1 PE = KE + W_{\text{friction}} Conservation of Energy
2 -\frac{1}{2}mv^2 + mgh = F_{\text{friction}}d Plug in formulas for each type of energy and solve for Fd (work)
3 F_{\text{friction}} = \mu mg\cos(\theta) Friction force is the product of the coefficient of friction, normal force, and gravity.
4 -\frac{1}{2}mv^2 + mgh = \mu mg\cos(\theta)d Substitute friction force from Step 3.
5 sin(\theta) = h/d Solve for h in terms of d and plug into the equation above
5 v = \sqrt{2g(dsin\theta) + 2\mu g\cos(\theta)s} Rearrange and solve for v.

The snowboarder’s speed at the base of the incline using energy considerations is \boxed{~25.5 , \text{m/s}}.

Now, let’s use this speed to calculate the distance traveled on the level surface.

Step Formula Derivation Reasoning
1 W_{\text{friction}} = F_{\text{friction}} \cdot d Work done by friction is the product of friction force and distance (d).
2 KE = W_{\text{friction}} Conservation of Energy
3 \frac{1}{2}mv^2 = \mu mgd Substitute KE expression W.
4 d = \frac{\frac{1}{2}mv^2}{\mu mg} Solve for d.

The distance the snowboarder travels along the level surface, calculated using energy considerations is \boxed{267.60 , \text{m}}.

  1. ~ 25.1 m/s
  2.  ~ 267.60 m

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  1. ~ 25.1 m/s
  2.  ~ 267.60 m

Nerd Notes

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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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