Objective: Find the shortest time in which the person can be brought out of the cave without breaking the rope.
Step | Formula Derivation | Reasoning |
---|---|---|
1 | F_{\text{net}} = T – w | The net force is the tension minus the weight of the person. |
2 | F_{\text{net}} = ma | Newton’s second law, net force equals mass times acceleration. |
3 | T = w + ma | Rewrite the net force equation to solve for tension. |
4 | a = \frac{T – w}{m} | Isolate acceleration, a. |
5 | m = \frac{w}{g} | The mass of the person is weight divided by gravitational acceleration. |
6 | a = \frac{T – w}{w/g} | Substitute m with w/g into the acceleration equation. |
7 | a = \frac{g(T – w)}{w} | Rearrange the equation for a. |
8 | a = \frac{9.8\text{ m/s}^2(592\text{ N} – 4.92 \times 10^2\text{ N})}{4.92 \times 10^2\text{ N}} | Substitute the values of T, w, and g. |
9 | a = \frac{9.8\text{ m/s}^2(592\text{ N} – 492\text{ N})}{492\text{ N}} | Calculate the values in the parentheses. |
10 | a = \frac{9.8\text{ m/s}^2 \times 100\text{ N}}{492\text{ N}} | Simplify the numerator. |
11 | a = 2\text{ m/s}^2 | Calculate the acceleration. |
For the Shortest Time
Step | Formula Derivation | Reasoning |
---|---|---|
1 | d = \frac{1}{2}at^2 | Kinematic equation for distance with initial velocity zero. |
2 | t = \sqrt{\frac{2d}{a}} | Solve for time, t. |
3 | t = \sqrt{\frac{2 \times 35.2\text{ m}}{2\text{ m/s}^2}} | Substitute the values of d and a. |
4 | t = \sqrt{\frac{70.4\text{ m}}{2\text{ m/s}^2}} | Multiply the numerator. |
5 | t = \sqrt{35.2\text{ s}^2} | Divide the numerator by the acceleration. |
6 | t = 5.934\text{ s} | Take the square root to find time. |
Final answer for the shortest time: \boxed{t = 5.934\text{ s}}
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A car starts from rest and accelerates uniformly over a time of 5 seconds for a distance of 100 m. Determine the acceleration of the car.
Traveling at a speed of 15.9 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.659. What is the speed of the automobile after 1.59 s have elapsed? Ignore the effects of air resistance.
The steepest street in the world is Baldwin Street in Dunedin, New Zealand. It has an inclination angle of 38.0° with respect to the horizontal. Suppose a wooden crate with a mass of 25.0 kg is placed on Baldwin Street. An additional force of 59 N must be applied to the crate perpendicular to the pavement in order to hold the crate in place. If the coefficient of static friction between the crate and the pavement is 0.599, what is the magnitude of the frictional force?
You stand at the edge of a vertical cliff and throws a stone vertically upwards. The stone leaves your hand with a speed v = 8.0 m/s. The time between the stone leaving your hand and hitting the sea is 3.0 s. Assume air resistance is negligible. Calculate:
Two students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down.
t = 5.94 s
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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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