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Step Derivation/Formula Reasoning
1 KE_{\text{initial}} = \frac{1}{2} m v^2 Calculate the initial kinetic energy of the vehicle using its speed before it halts. Here v = 12.3 \text{ m/s}.
2 W_{\text{friction}} = f_k \cdot d = \mu_k \cdot m \cdot g \cdot \cos(\theta) \cdot d The work done by friction, where \mu_k = 0.650 is the coefficient of kinetic friction, g = 9.8 \text{ m/s}^2 is acceleration due to gravity, \theta = 18^\circ, and d is the distance the vehicle slides.
3 W_{\text{gravity}} = m \cdot g \cdot \sin(\theta) \cdot d The work done by gravity while the vehicle moves down the incline.
4 KE_{\text{initial}} = W_{\text{friction}} + W_{\text{gravity}} By the work-energy principle, the initial kinetic energy is converted into work done against friction plus the work done by gravity.
5 \frac{1}{2} m v^2 = \mu_k mg\cos(\theta)d + mg\sin(\theta)d Substitute expressions from steps 1, 2, and 3 into the work-energy equation.
6 d = \frac{\frac{1}{2} v^2}{\mu_k g \cos(\theta) + g \sin(\theta)} Solve for d, distance the vehicle slides. Notice that mass m cancels out.
7 d = \frac{\frac{1}{2} (12.3)^2}{0.650 \times 9.8 \times \cos(18^\circ) + 9.8 \times \sin(18^\circ)} Substitute numerical values for v, \mu_k, g, and \theta to find the value of d that represents the distance the vehicle slides until it stops.
8 d  = 8.32 \,\text{m} Calculated value.

To solve part b, look at the equation derived in step 6 of part a. Notice that the angle and distance traveled up the incline are inversely proportional. This means the greater the angle \theta the shorter the distance traveled.

Step Derivation/Formula Reasoning
1 d' = \frac{\frac{1}{2} v^2}{\mu_k g \cos(27^\circ) + g \sin(27^\circ)} Recalculate the distance with the increased angle of 1.5 \times 18^\circ = 27^\circ.
2 d' = \frac{d}{\cos(27^\circ) + \tan(27^\circ) \cdot \sin(27^\circ)} Using the previous formula of d, we express the new sliding distance d' in terms of the old distance d.
3 d' = 7.41 \, \text{meters} At an angle of 27^\circ the vehicale would slide up 7.41 meters, which is less than the orginal dsitance of 8.32 meters.
3 \frac{d'}{d} The ratio \frac{d'}{d} shows how much further the vehicle would slide relative to d.
4 \frac{7.41}{8.32} = .89 Thus the new distance d' = .89d

Part (c):

Step Derivation/Formula Reasoning
1 W_{\text{gravity,up}} = -m \cdot g \cdot \sin(\theta) \cdot d The work done by gravity as the vehicle slides up the incline is negative since gravity opposes the motion.
2 \frac{1}{2} m v^2 = \mu_k m g \cos(\theta) d – m g \sin(\theta) d Work-energy principle applied while moving up. The kinetic energy has to overcome both friction and an upward gravity force.
3 d_{\text{up}} = \frac{\frac{1}{2} v^2}{\mu_k g \cos(\theta) – g \sin(\theta)} Solve for the distance the vehicle would slide up the incline.
4 d_{\text{up}} < d The distance d_{\text{up}} will be lesser than d since gravity now acts against the motion, reducing the sliding distance relative to sliding down.

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1. d = 8.32 \,\text{meters}
2. d' = .89d

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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)}
\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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