^{2} 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?

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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- Statistics

Intermediate

Mathematical

GQ

Two objects (49.0 and 24.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley hangs from the ceiling. Find the acceleration of the objects and the tension in the string.

- Atwood Machine, Linear Forces

Advanced

Mathematical

MCQ

An object undergoes constant acceleration. Starting from rest, the object travels 5 m in the first second. Then it travels 15 meters in the next second. What additional distance will be covered in the third second?

- 1D Kinematics

Advanced

Mathematical

GQ

A space probe far from the Earth is traveling at 14.8 km/s. It has mass 1312 kg. The probe fires its rockets to give a constant thrust of 156 kN for 220 seconds. It accelerates in the same direction as its initial velocity. In this time it burns 150 kg of fuel. Calculate final speed of the space probe in km/s.

*Note: This is a bonus question. Skip if you haven’t yet taken calculus. *

- Linear Forces, Momentum

Beginner

Conceptual

MCQ

The displacement x of an object moving in one dimension is shown above as a function of time t. The acceleration of this object must be

- 1D Kinematics, Motion Graphs

Intermediate

Mathematical

FRQ

A 0.035 kg bullet moving horizontally at 350 m/s embeds itself into an initially stationary 0.55 kg block. Air resistance is negligible.

- Energy, Linear Forces, Momentum

Intermediate

Mathematical

MCQ

A force *F* is used to hold a block of mass *m* on an incline as shown in the diagram above. The plane makes an angle of \theta with the horizontal and *F* is perpendicular to the plane. The coefficient of friction between the plane and the block is *µ*. What is the minimum force, *F*, necessary to keep the block at rest?

- Inclines, Linear Forces

Advanced

Conceptual

FRQ

A student is watching their hockey puck slide up and down an incline. They give the puck a quick push along a frictionless table, and it slides up a 30° rough incline (µ_{k} = .4) of distance *d*, with an initial speed of 5 m/s, and then it slides back down.

- Inclines, Linear Forces

Intermediate

Mathematical

FRQ

A rescue helicopter lifts a 79 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s^{2} and is lifted through a distance of 11 m.

- Energy, Linear Forces

Advanced

Mathematical

FRQ

*A* has a mass of 3.2 kg and block *B* a mass of 2.4 kg. The pulley is frictionless and has no mass.

- Atwood Machine, Linear Forces

Intermediate

Mathematical

GQ

- Inclines, Linear Forces

*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

Conversion Example

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 |

- Some answers may be slightly off by 1% depending on rounding, etc.
- Answers will use different values of gravity. Some answers use 9.81 m/s
^{2}, and other 10 m/s^{2 }for calculations. - 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 .
- Bookmark questions that you can’t solve so you can come back to them later.
- 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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