# A person whose weight is 4.92 × 102 N is being pulled up vertically by a rope from the bottom of a cave that is 35.2 m deep. The maximum tension that the rope can withstand without breaking is 592 N. What is the shortest time, starting from rest, in which the person can be brought out of the cave?

t = 5.94 s

0

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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t = 5.94 s

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##### Made By Nerd-Notes.com
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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