Objective: Determine whether it takes longer for the hockey puck to slide up or down a distance d on an incline with kinetic friction μk=0.4, 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 =0 m/s at the top of incline. |
3 | t_{\text{up}} = \frac{v – u}{a_{\text{up}}} | Time taken tup is found using the kinematic equation for time. |
4 | Substitute and solve for tup. 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 tdown using the kinematic equation for constant acceleration. |
7 | Substitute and solve for tdown. 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 d 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 d 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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The coefficient of kinetic friction is
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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Kinematics | Forces |
---|---|
\Delta x = v_i t + \frac{1}{2} at^2 | F = ma |
v = v_i + at | F_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 Motion | Energy |
---|---|
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 |
Momentum | Torque and Rotations |
---|---|
p = m v | \tau = r \cdot F \cdot \sin(\theta) |
J = \Delta p | I = \sum mr^2 |
p_i = p_f | L = I \cdot \omega |
Simple Harmonic Motion |
---|
F = -k x |
T = 2\pi \sqrt{\frac{l}{g}} |
T = 2\pi \sqrt{\frac{m}{k}} |
Constant | Description |
---|---|
g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
\mu_k and \mu_s | Coefficients 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. |
k | Spring 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 |
Variable | SI 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)} |
Variable | Derived 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.
Start with the given measurement: \text{5 km}
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}}
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}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|
Pico- | p | 10^{-12} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
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