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Objective: Determine whether it takes longer for the hockey puck to slide up or down a distance on an incline with kinetic friction , given an initial speed of 5 m/s.

Analysis for Upward Motion:

Step Formula Derivation Reasoning
1 a_{\text{up}} = -g(\sin(\theta) + \mu_k \cos(\theta)) Acceleration is due to gravity and kinetic friction opposing the motion.
2 v^2 = u^2 + 2a_{\text{up}}d Use the kinematic equation for final velocity. v m/s at the top of incline.
3 t_{\text{up}} = \frac{v – u}{a_{\text{up}}} Time taken is found using the kinematic equation for time.
4 Substitute and solve for . u = 5 \text{ m/s}, \theta = 30^\circ, \mu_k = 0.4 Calculate time up.

Analysis for Downward Motion:

Step Formula Derivation Reasoning
5 a_{\text{down}} = g(\sin(\theta) – \mu_k \cos(\theta)) Acceleration is due to gravity assisted by friction.
6 t_{\text{down}} = \sqrt{\frac{2d}{a_{\text{down}}}} Time taken using the kinematic equation for constant acceleration.
7 Substitute and solve for u = 0 \text{ m/s} at the peak, \theta = 30^\circ, \mu_k = 0.4 Calculate time down.

Calculating the time for both upward and downward motions:

Step Result
8 t_{\text{up}} \approx 0.60 \text{ s}
9 t_{\text{down}} \approx 1.15 \text{ s}

The time taken for the hockey puck to move up the distance is approximately 0.60 seconds, while the time to slide down the same distance is approximately 1.15 seconds. Therefore, it takes longer for the puck to move down the distance than to move up.

This result may seem counterintuitive, but it’s important to note that the initial conditions (starting speed and angle of incline) and the presence of friction significantly influence the motion. The initial upward speed allows the puck to cover the upward distance quickly, while friction continuously slows it down during both upward and downward motion, affecting the total time for each path.

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It takes less time to slide up. This is because there is a greater net force on the way up (gravity and friction work together to slow the block down quicker).

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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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