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Step Derivation/Formula Reasoning
1 E_{\text{spring}} = \frac{1}{2} k x^2 The initial energy stored in the spring is given by the spring potential energy formula, where k is the spring constant and x is the compression.
2 E_{\text{spring}} = \frac{1}{2} \times 1800 \times 0.3^2 Substitute k = 1800 \, \text{N/m} and x = 0.3 \, \text{m} into the spring energy formula to calculate the initial energy.
3 E_{\text{spring}} = 81 \, \text{J} Calculate the total energy stored in the spring.
4 W_f = f_k \times d Calculate the work done by friction, where f_k is the kinetic friction force and d is the distance over which the force acts.
5 f_k = \mu_k \times m \times g Kinetic friction force is the coefficient of kinetic friction times the normal force. Here, \mu_k = 0.12 , m = 6 \, \text{kg} , and g = 9.8 \, \text{m/s}^2 . On horizontal surface, normal force equals mg.
6 f_k = 0.12 \times 6 \times 9.8 = 7.056 \, \text{N} Calculate the kinetic friction force on the horizontal surface.
7 W_f = 7.056 \times 3.3 = 23.28 \, \text{J} Calculate the work done by friction over the 3 m horizontal surface and the distance it slides while the spring decompresses (.3 m) for a total distance of 3.3 m.
8 W_{f, \text{slope}} = f_{k, \text{slope}} \times d_{\text{slope}} Work done by friction on the slope, where f_{k, \text{slope}} is the kinetic friction force on the slope and d_{\text{slope}} = 4.5 \, \text{m} .
9 f_{k, \text{slope}} = \mu_{k, \text{slope}} \times m \times g \times \cos(\theta) On the slope, the normal force is reduced by the cosine of the slope angle \theta = 12^\circ .
10 W_{f, \text{slope}} = \mu_{k, \text{slope}} \times m \times g \times \cos(\theta) \times d_{\text{slope}} Substitute equation from step 9 into the equation for work done by the friction of the slope (step 8).
11 W_{f, \text{slope}} = \mu_{k, \text{slope}} \times 258.818 \, \text{J} Plug in given values and simplify the equation for work done by the friction of the slope as much as possible.
12 PE_{\text{top}} = mgh = 5 \times 9.8 \times 4.5sin(12^\circ) \approx 55.01 \, \text{J} Now find the final potential energy at the top of the incline.
13 E_{\text{spring}} = W_f + W_{f, \text{slope}} + PE_{\text{top}} Finally add up all the energy types together. Use the law of conservation of energy which tells us the initial energy from the spring will turn into work done by friction on level ground, work done by friction on the slope, and the potential energy gained traveling up the slope.
14 81 = 23.28 + (\mu_{k, \text{slope}} \times 258.818) + 55.01 Substitute all numbers found in the previous steps.
15 \boxed{\mu_{k, \text{slope}} \approx .01} Solve for the coefficient of kinetic friction on the slope.

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\mu_{k, \text{slope}} \approx .01

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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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