/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 In 2.0 minutes, a ski lift raise... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In 2.0 minutes, a ski lift raises four skiers at constant speed to a height of \(140 \mathrm{m}\). The average mass of each skier is \(65 \mathrm{kg}\). What is the average power provided by the tension in the cable pulling the lift?

Short Answer

Expert verified
The average power is approximately 2.97 kW.

Step by step solution

01

Calculate Total Weight

The total weight of the skiers is the sum of their individual weights. Since each skier has a mass of 65 kg, the total mass is \( 4 \times 65 \text{ kg} = 260 \text{ kg} \). The total weight is given by \( W = mg \), where \( g = 9.8 \text{ m/s}^2 \). Therefore, \( W = 260 \times 9.8 = 2548 \text{ N} \).
02

Determine Work Done

Work done is calculated using \( W = Fd \), where \( F \) is the force (which is equal to the weight), and \( d \) is the height. Here, \( F = 2548 \text{ N} \) and \( d = 140 \text{ m} \). Thus, \( W = 2548 \times 140 = 356720 \text{ J} \).
03

Calculate Average Power

Power is the rate of doing work, given by \( P = \frac{W}{t} \), where \( t \) is the time in seconds. Since the time is 2 minutes, convert it to seconds: \( t = 2 \times 60 = 120 \text{ s} \). Therefore,\( P = \frac{356720}{120} = 2972.67 \text{ W} \) or approximately \( 2.97 \text{ kW} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ski Lift
A ski lift is a transportation system used primarily in ski resorts to move skiers from the bottom of a hill to the top efficiently and comfortably. It uses a motorized cable to pull chairs or gondolas that carry people up a slope. The lift operates at a constant speed, making it a reliable method of transport for both leisurely rides and competitive skiing events.
For anyone involved in winter sports, understanding how a ski lift operates can enhance their experience and safety. A ski lift consists of several important components:
  • The cable, which is driven by a motor.
  • The pulleys and wheels that guide the cable through the lift.
  • The chairs or cabins that carry skiers.
  • Safety and emergency systems to protect passengers in case of malfunction.
While the mechanics of the ski lift itself might seem simple, the physics behind its operation, involving average power and energy calculations, can be quite interesting.
Work Done
In physics, "work done" is a measure of energy transfer that occurs when a force acts over a distance. For our ski lift scenario, this involves lifting the skiers to a new height.
The formula used to calculate work done is given by:
  • done is \( W = F \times d \) where \( F \) is the force applied, and \( d \) is the distance over which the force is applied.
  • Force, in this case, is equivalent to the weight of the skiers due to gravity, calculated as \( F = mg \) where \( m \) is the total mass and \( g \) is the gravitational acceleration (approximately 9.8 m/s²).
  • For example, if the force is 2548 N and the height is 140 m, then the work done is \( W = 2548 \times 140 = 356720 \text{ J} \).
This energy transfer helps skiers reach the top of the slope by overcoming gravitational potential energy. The calculation of work done helps us quantify the energy required for the ski lift's operation, ensuring effective and efficient performance.
Forces
Forces play a crucial role in the operation of ski lifts. A force is any interaction that, when unopposed, will change the motion of an object. In our ski lift example, the primary force is the weight of the skiers, which is countered by the tension in the lift's cable pulling them upwards.
Some key aspects to remember about forces in this context include:
  • The weight of the skiers is calculated using \( W = mg \).
  • The tension force in the cable is equal in magnitude but opposite in direction to the weight of the skiers when moving at constant speed.
  • Normal forces can affect the stability and comfort of the lift ride.
Understanding these forces helps in designing and operating the ski lift safely and efficiently. Engineers use this knowledge to maintain balance and tension in the lift system to ensure smooth and constant transport to the top of the hill.
Energy Calculations
Energy calculations help us understand how much energy is required to perform a task, such as lifting skiers up a mountain. In this context, we calculate the average power output necessary to keep the ski lift functioning efficiently.
The concept of energy calculations involves understanding both potential and kinetic energy:
  • Potential energy is related to the height being climbed. It's the energy stored due to the object's position and is given by \( PE = mgh \).
  • Kinetic energy isn't primarily a factor here because the ski lift moves at a constant speed; however, it initially accounts for getting the cable system up to its operating speed.
  • Average power, critical for operating the ski lift, is the rate at which work is done. It's calculated by dividing the work by time, \( P = \frac{W}{t} \).
For instance, lifting a total mass of skiers to a height of 140 meters in 2 minutes results in an average power of approximately 2972.67 Watts. This output ensures the lift operates smoothly, providing the energy needed to overcome gravitational forces and deliver consistent performance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

As a sailboat sails 52 m due north, a breeze exerts a constant force \(\overrightarrow{\mathbf{F}}_{1}\) on the boat's sails. This force is directed at an angle west of due north. A force \(\overrightarrow{\mathbf{F}}_{2}\) of the same magnitude directed due north would do the same amount of work on the sailboat over a distance of just \(47 \mathrm{m} .\) What is the angle between the direction of the force \(\overrightarrow{\mathbf{F}}_{1}\) and due north?

A Makeshift Elevator. While exploring an elaborate tunnel system, you and your team get lost and find yourselves at the bottom of a \(450-\mathrm{m}\) vertical shaft. Suspended from a thick rope (near the floor) is a large rectangular bucket that looks like it had been used to transport tools and debris up and down the tunnel. Mounted on the floor near one of the walls is a gasoline engine \((3.5 \mathrm{hp})\) that turns a pulley and rope, and a sign that reads "Emergency Lift." It is clear that the engine is used to drive the bucket up the shaft. On the wall next to the engine is a sign indicating that a full tank of gas will last exactly 15 minutes when the engine is running at full power. You open the engine's gas tank and estimate that it is \(1 / 4\) full, and there are no other sources of gasoline. (a) Assuming zero friction, if you send your team's lightest member (who weighs \(125 \mathrm{lb}\) ), and the bucket weighs \(150 \mathrm{lb}\) when empty, how far up the shaft will the engine take her (and the bucket)? Will it get her out of the mine? (b) Assuming an effective collective friction (from the pulleys, etc.) of \(\mu_{\mathrm{eff}}=0.10\) (so that \(F_{\mathrm{f}}=\mu_{\mathrm{eff}} M g,\) where \(M\) is the total mass of the bucket plus team member), will the engine (with a \(1 / 4\) -full tank of gas) lift her to the top of the shaft?

A \(16-\mathrm{kg}\) sled is being pulled along the horizontal snow-covered ground by a horizontal force of \(24 \mathrm{N}\). Starting from rest, the sled attains a speed of \(2.0 \mathrm{m} / \mathrm{s}\) in \(8.0 \mathrm{m} .\) Find the coefficient of kinetic friction between the runners of the sled and the snow.

? 67.0-kg person jumps from rest off a 3.00-m-high tower straight down into the water. Neglect air resistance. She comes to rest \(1.10 \mathrm{m}\) under the surface of the water. Determine the magnitude of the average force that the water exerts on the diver. This force is nonconservative.

Juggles and Bangles are clowns. Juggles stands on one end of a teeter-totter at rest on the ground. Bangles jumps off a platform \(2.5 \mathrm{m}\) above the ground and lands on the other end of the teeter-totter, launching Juggles into the air. Juggles rises to a height of \(3.3 \mathrm{m}\) above the ground, at which point he has the same amount of gravitational potential energy as Bangles had before he jumped, assuming both potential energies are measured using the ground as the reference level. Bangles' mass is \(86 \mathrm{kg}\). What is Juggles' mass?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.