/*! 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 69 If a ski lift raises 100 passeng... [FREE SOLUTION] | 91Ó°ÊÓ

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If a ski lift raises 100 passengers averaging \(660 \mathrm{~N}\) in weight to a height of \(150 \mathrm{~m}\) in \(60.0 \mathrm{~s}\), at constant speed, what average power is required of the force making the lift?

Short Answer

Expert verified
The average power required is 165 kW.

Step by step solution

01

Understand the Concept of Power

Power is defined as the rate at which work is done. In this problem, the ski lift raises passengers to a height, therefore doing work against gravity. We use the formula for power, \[ P = \frac{W}{t} \]where \(P\) is power, \(W\) is work done, and \(t\) is the time.
02

Calculate Total Work Done

The work done is equal to the change in gravitational potential energy, given by \[ W = mgh \]where \(m\) is the total mass of the passengers, \(g = 9.8\, \mathrm{m/s^2}\) is the acceleration due to gravity, and \(h = 150\, \mathrm{m}\) is the height. First, calculate the total weight (force due to gravity) of the passengers as \[ F = 100 \times 660 = 66000 \, \mathrm{N} \].Since the work done \(W\) is \(F \times h\),\[ W = 66000 \times 150\, = 9900000 \, \mathrm{J} \].
03

Calculate Average Power

With the work done calculated, use the power formula:\[ P = \frac{9900000}{60} \, \mathrm{W} \].This calculates to \[ P = 165000 \, \mathrm{W} \] or 165 kW.

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Key Concepts

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

Work and Energy
Work and Energy are fundamental concepts in physics. Work is defined as the energy transferred to or from an object by a force acting on it over a distance. In simpler terms, if you push or pull something and it moves, you are doing work. The formula to calculate work is:
  • \[ W = F imes d \]
where \(W\) is work, \(F\) is the force applied, and \(d\) is the distance over which the force is applied. If the direction of force and movement differ, use the component of the force in the direction of movement.
Energy, on the other hand, is the ability to do work. It exists in many forms but, in this scenario, we primarily focus on gravitational potential energy. Understanding the relationship between work and energy is essential when solving physics problems as energy changes can describe the work done on or by an object. In our ski lift example, the work done by the lift is converting energy from one form to another, specifically overcoming gravitational force.
Gravitational Potential Energy
Gravitational Potential Energy (GPE) is the energy that an object possesses because of its position in a gravitational field. The higher an object is lifted against gravity, the more GPE it has. You can calculate this energy using the formula:
  • \[ U = mgh \]
where \(U\) is the gravitational potential energy, \(m\) is the mass, \(g\) is the gravitational acceleration \((9.8 \, \mathrm{m/s^2})\), and \(h\) is the height above the reference point.
In our ski lift scenario, all passengers are lifted to a height, adding to their gravitational potential energy. Calculating GPE helps us determine the amount of work needed to move something against gravity. Hence, when the ski lift raises passengers to a height of 150 meters, it increases their GPE by a specific amount, calculated as work done against gravity.
Force and Motion
Force and Motion are closely linked concepts in physics explored through Newton's laws of motion. Force is any interaction that changes the motion of an object, and it is calculated using Newton's second law:
  • \[ F = ma \]
where \(F\) is force, \(m\) is mass, and \(a\) is acceleration. In our example, the force exerted by the ski lift is a necessary component to counteract gravity and move passengers to a higher point.
Motion occurs when an object changes its position over time due to a force. Although the ski lift moves at a constant speed, work is still done to lift against gravity. This demonstrates how force operates even when motion is constant, emphasizing the importance of understanding both concepts when examining problems involving lifting or moving objects at constant speeds.

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Most popular questions from this chapter

The only force acting on a \(2.0 \mathrm{~kg}\) canister that is moving in an \(x y\) plane has a magnitude of \(5.0 \mathrm{~N}\). The canister initially has a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) direction and some time later has a velocity of \(6.0 \mathrm{~m} / \mathrm{s}\) in the positive \(y\) direction. How much work is done on the canister by the \(5.0 \mathrm{~N}\) force during this time?

A helicopter lifts a \(72 \mathrm{~kg}\) astronaut \(15 \mathrm{~m}\) vertically from the ocean by means of a cable. The acceleration of the astronaut is \(g / 10 .\) How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her? Just before she reaches the helicopter, what are her (c) kinetic energy and (d) speed?

A fully loaded, slow-moving freight elevator has a cab with a total mass of \(1200 \mathrm{~kg}\), which is required to travel upward \(54 \mathrm{~m}\) in \(3.0\) min, starting and ending at rest. The elevator's counterweight has a mass of only \(950 \mathrm{~kg}\), and so the elevator motor must help. What average power is required of the force the motor exerts on the cab via the cable?

A \(45 \mathrm{~kg}\) block of ice slides down a frictionless incline \(1.5 \mathrm{~m}\) long and \(0.91 \mathrm{~m}\) high. A worker pushes up against the ice, parallel to the incline, so that the block slides down at constant speed. (a) Find the magnitude of the worker's force. How much work is done on the block by (b) the worker's force, (c) the gravitational force on the block, (d) the normal force on the block from the surface of the incline, and (e) the net force on the block?

An iceboat is at rest on a frictionless frozen lake when a sudden wind exerts a constant force of \(200 \mathrm{~N}\), toward the east, on the boat. Due to the angle of the sail, the wind causes the boat to slide in a straight line for a distance of \(8.0 \mathrm{~m}\) in a direction \(20^{\circ}\) north of east. What is the kinetic energy of the iceboat at the end of that \(8.0 \mathrm{~m} ?\)

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