Step | Derivation/Formula | Reasoning |
---|---|---|

1 | h = s \sin(\theta) | To find the vertical height h gained by the skier on the slope (where s = 695 meters, \theta = 34^\circ), we use the sine component of the inclined angle because height is opposed to gravity and directly vertical. |

2 | h = 695 \sin(34^\circ) | Calculate h by substituting the values of s and \theta. The sine of 34^\circ (from calculator or trigonometric table) is used to obtain h. |

3 | PE = mgh | Calculate the potential energy (PE) needed to lift a skier to height h. Here, m is mass (72 kg), g is the acceleration due to gravity (approximately 9.81 \, \text{m/s}^2), and h is the height obtained in the previous step. |

4 | PE = 72 \cdot 9.81 \cdot 695 \sin(34^\circ) | Substitute the values into the potential energy formula to find the energy required to transport one skier to the top of the slope. |

5 | P = \frac{PE \cdot R}{t} | Calculate the power (P) needed to transport R riders per minute. t is the time in seconds; for one minute, t = 60 seconds. |

6 | P = \frac{72 \cdot 9.81 \cdot 695 \sin(34^\circ) \cdot 5}{60} | Substitute the values to find the power required for ferrying 5 riders per minute. |

7 | P_{\text{total}} = \frac{P}{\text{efficiency}} | To find the total average power (P_{\text{total}}) supplied by the motor, we adjust for the efficiency of the ski lift, which only uses 65% of the energy supplied to overcome work against friction. |

8 | P_{\text{total}} = \frac{72 \cdot 9.81 \cdot 695 \sin(34^\circ) \cdot 5}{60 \cdot 0.65} | Calculate P_{\text{total}} by considering the efficiency. This power value indicates the power that must be supplied by the motor. |

9 | P_{\text{total}} \approx 35,192.76 \, \text{Watt} | Complete the calculations to get the final answer in Watts (rounded to sensible precision). |

Phy can also check your working. Just snap a picture!

- Statistics

Intermediate

Conceptual

MCQ

Two masses m_{1} and 4m_{1} are on an incline. Both surfaces have the same coefficient of kinetic friction. Both objects start from rest, at the same height. Which mass has the largest speed at the bottom?

- Energy

Advanced

Proportional Analysis

MCQ

A rocket of mass m is launched with kinetic energy *K _{0}*, from the surface of the Earth. How much less kinetic energy does the rocket have at an altitude of two Earth radii in terms of the gravitational constant,

- Energy

Intermediate

Conceptual

MCQ

You kick a ball straight up. Compare the sign of the work done by gravity on the ball while it goes up with the sign of the work done by gravity while it goes down.

- Energy

Intermediate

Mathematical

MCQ

A child pushes horizontally on a box of mass *m* with constant speed *v* across a rough horizontal floor. The coefficient of friction between the box and the floor is *µ*. At what rate does the child do work on the box?

- Energy

Intermediate

Conceptual

MCQ

A rubber ball bounces off of a wall with an initial speed v and reverses its direction so its speed is v right after the bounce. As a result of this bounce, which of the following quantities of the ball are conserved?

- Energy, Momentum

35,192.76 Watts

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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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