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An elevator has mass \(600 \mathrm{~kg},\) not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of \(20.0 \mathrm{~m}\) (five floors) in \(16.0 \mathrm{~s}\), and it is driven by a motor that can provide up to 40 hp to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has mass \(65.0 \mathrm{~kg}\).

Short Answer

Expert verified
The maximum number of passengers that can ride in the elevator is about \(N_{passengers}\) (to the nearest whole number, as we don't consider fractions of a person).

Step by step solution

01

Compute the work done

First, calculate the work done to lift the elevator. The work done against gravity is given by the formula: \( Work = mass \times gravity \times height \). The elevator, not including passengers, has a mass of 600 kg, the acceleration due to gravity is \(9.8 m/s^2\), and the vertical distance is 20 m. So, the work done for the elevator is \( Work_{elevator} = 600 kg \times 9.8 m/s^2 \times 20 m \).
02

Compute the power required

Next, compute the power required to lift the elevator. Power is the work done per unit time. The time taken to lift the elevator is given as 16.0 s. Calculate the power using the formula: \( Power = Work/Time \). Therefore, the power required for the elevator is \( Power_{elevator} = Work_{elevator}/16 s \).
03

Compute the power available

The power provided by the motor is given as 40 hp. To use this information, it needs to be converted to the standard unit of power, watts (W). 1 hp equals 746 W, so the total power provided by the motor is \( Power_{total} = 40 hp \times 746 W/hp \).
04

Compute the power available for passengers

The motor's power is used both to lift the elevator and the passengers. Thus, we need to subtract the power used by the elevator from the total power provided by the motor to find the remaining power for the passengers. This is given by: \( Power_{passenger} = Power_{total} - Power_{elevator} \).
05

Compute maximum passengers

Finally, we will compute the maximum number of passengers that can ride in the elevator. The work done per passenger is given by the formula: \( Work_{passenger} = mass_{passenger} \times gravity \times height \), where mass of a passenger is given as 65 kg. The power required per passenger is computed in the same way as in step 2: \( Power_{required per passenger} = Work_{passenger}/16 s \). The maximum number of passengers is then given by: \( N_{passengers} = Power_{passenger} / Power_{required per passenger} \).

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Key Concepts

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

Elevator Physics
Elevator physics involves understanding how elevators move vertically and how this movement requires energy to counteract gravitational forces. At the heart of this is the principle of work done against gravity. When an elevator moves up, it must exert force equal to the weight of the elevator and its contents to achieve its designated height. This situation involves a delicate balance of mass, gravitational acceleration, and energy produced by a motor.

In the problem, we're looking at an elevator with a specific mass and height to be reached within a set time frame. This scenario is typical of real-world applications where elevators must adhere to strict performance standards, combining physics fundamentals with engineering constraints. Understanding these dynamics helps in designing efficient elevator systems, capable of safely transporting passengers while optimizing energy use.
Work and Power
Work and power are fundamental concepts in physics that describe how force is applied over distances and the rate at which this work is done, respectively. Work, denoted by the formula \[ \text{Work} = \text{mass} \times \text{gravity} \times \text{height} \]measures the total energy required to move an object up against gravity. Power calculates how fast this energy is used, and it's given by \[ \text{Power} = \frac{\text{Work}}{\text{time}} \].

In our elevator problem, the calculated work represents the energy needed to lift the elevator without passengers. This is dependent on lifting the entire mass of the elevator up to a given height, battling gravitational forces. Then, with the work known, the power required to achieve this in the time provided (16 seconds) is found. This power represents the effectiveness of the motor to provide continuous motion at a constant speed. In real scenarios, understanding the work and power requirements ensures that motors are appropriately sized and energy is used efficiently.
Gravitational Force
Gravitational force is the natural phenomenon by which objects with mass attract one another. Here on Earth, it gives weight to physical objects and is a crucial factor in the operation of elevators. The force acting on an object due to gravity is calculated using \[ \text{Force} = \text{mass} \times \text{gravity} \].

In the problem, the gravitational force determines how much work is needed to raise both the elevator and its passengers. With Earth's gravitational acceleration approximated as 9.8 m/s², the force is proportional to both the mass of the elevator and that of the passengers. Elevators must overcome this force to ascend, which is why understanding it is vital for calculating required energies and selecting suitable motors for elevator operation.
Energy Efficiency
Energy efficiency in the context of elevators refers to optimizing how much energy is used relative to how much work is done. An efficient elevator will perform the necessary work—lifting its load using the least possible power, which is particularly important in building design and engineering.

In this exercise, the motor's efficiency is implied by comparing its capacity (measured in horsepower converted to watts) against the required power for the task. The calculation focuses on ensuring that the motor is not only adequate for lifting the elevator but also leaves additional capacity for passengers. By finding the balance between the motor's rated power and the work done, the elevator's power use is optimized. This efficiency minimizes energy waste, reduces operational costs, and supports sustainability efforts. Keeping such systems effective requires ongoing assessment of mechanical performance and power requirements, ensuring they meet or exceed their design specifications.

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Most popular questions from this chapter

You are a member of an Alpine Rescue Team. You must project a box of supplies up an incline of constant slope angle \(\alpha\) so that it reaches a stranded skier who is a vertical distance \(h\) above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient \(\mu_{\mathrm{k}}\). Use the work-energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the skier. Express your answer in terms of \(g, h, \mu_{k},\) and \(\alpha\)

A spring of force constant \(300.0 \mathrm{~N} / \mathrm{m}\) and unstretched length \(0.240 \mathrm{~m}\) is stretched by two forces, pulling in opposite directions at opposite ends of the spring, that increase to \(15.0 \mathrm{~N}\). How long will the spring now be, and how much work was required to stretch it that distance?

A net horizontal force \(F\) is applied to a box with mass \(M\) that is on a horizontal, frictionless surface. The box is initially at rest and then moves in the direction of the force. After the box has moved a dis- tance \(D,\) the work that the constant force has done on it is \(W_{D}\) and the speed of the box is \(V\). The equation \(P=F v\) tells us that the instanta neous rate at which \(F\) is doing work on the box depends on the speed of the box. (a) At the point in the motion of the box where the force has done half the total work, and so has done work \(W_{D} / 2\) on the box that started from rest, in terms of \(V\) what is the speed of the box? Is the speed at this point less than, equal to, or greater than half the final speed? (b) When the box has reached half its final speed, so its speed is \(V / 2,\) how much work has been done on the box? Express your answer in terms of \(W_{D}\). Is the amount of work done to produce this speed less than, equal to, or greater than half the work \(W_{D}\) done for the full displacement \(D ?\)

CP You are pushing a large box across a frictionless floor by applying a constant horizontal force. If the box starts at rest, you have to do work \(W_{1}\) in order for the box to travel a distance \(d\) in time \(t .\) How much work would you have to do, in terms of \(W_{1}\), to make the box go the same distance in half the time?

A \(75.0 \mathrm{~kg}\) painter climbs a ladder that is \(2.75 \mathrm{~m}\) long and leans against a vertical wall. The ladder makes a \(30.0^{\circ}\) angle with the wall. (a) How much work does gravity do on the painter? (b) Does the answer to part (a) depend on whether the painter climbs at constant speed or accelerates up the ladder?

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