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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 \(\mathrm{hp}\) 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 elevator can carry a maximum of 28 passengers.

Step by step solution

01

Calculate Work Done by the Elevator

The work done by the elevator can be calculated using the formula for work done against gravity, which is \( W = m \cdot g \cdot h \). First, we determine the total work that needs to be done by including the passengers. Assume \( n \) passengers each of mass 65.0 \( \mathrm{kg} \). The total mass \( m \) of the system (elevator plus passengers) is \( 600 + 65n \) \( \mathrm{kg} \). Hence, \( W = (600 + 65n) \cdot 9.8 \cdot 20.0 \).
02

Convert Motor Power from Horsepower to Watts

The maximum power of the motor is given in horsepower, thus needs to be converted to watts because power in physics is typically calculated in watts. 1 horse power \( = 746 \) watts. So, the power \( P \) in watts is \( 40 \times 746 = 29840 \) watts.
03

Calculate Time-adjusted Power

Knowing the elevator ascends in \( 16.0 \) seconds, we relate power to work over time using \( P = \frac{W}{t} \). Rewriting gives \( W = P \cdot t = 29840 \times 16.0 = 477440 \) joules. This is the maximum work the motor can do in that time period.
04

Equate and Solve for Passenger Number n

Now we set the work done by the motor equal to the work required to lift the elevator and passengers: \[(600 + 65n) \cdot 9.8 \cdot 20.0 = 477440\] Solving for \( n \): \[(600 + 65n) \cdot 196 = 477440\] Divide both sides by 196: \[600 + 65n = 2436.734693\] Finally, solve for \( n \) to find that \( 65n = 2436.734693 - 600 \). Then \[65n = 1836.734693\] Divide by 65: \[n = 28.26\].
05

Round Down for Whole Number of Passengers

Since \( n \) represents the number of passengers whole number values are needed. Thus, the maximum number of passengers can only be \( 28 \), as any fractional value means the elevator cannot take the additional passenger.

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Key Concepts

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

Work and Energy
In the study of physics, work and energy are closely tied to how forces are exerted and the movement they produce. Work, symbolized by the formula \( W = m \cdot g \cdot h \), indicates the energy transferred when a force moves an object over a distance. Here, \( m \) is the mass of the elevator plus passengers, \( g \) is the gravitational acceleration (approximately 9.8 \( \text{m/s}^2 \)), and \( h \) is the height change, 20 meters for the ascent.

To understand this, think about how lifting objects consumes energy. The greater the mass or the higher the object needs to be lifted, the more work is needed. This concept is crucial when determining the effort required by the elevator motor to overcome gravitational force and move upwards. Work done against gravity directly impacts the elevator system's efficiency and limits its passenger capacity, as excessive work demands more energy from the motor.

This relationship highlights why understanding energy transformations, especially in mechanical systems like elevators, is essential. Every joule of energy expended must be supplied adequately to maintain functional operation.
Power Conversion
Power conversion is critical when dealing with machines like elevators that convert energy into practical movement. Put simply, power is how quickly work is done, or energy is transferred. In this problem, the motor's power is initially given in horsepower, a non-standard unit in scientific calculations. To use it effectively in equations, it is converted into watts, the SI unit for power.

The conversion is as follows: 1 horsepower is equivalent to 746 watts, allowing the calculation of the motor's output as \( 40 \times 746 = 29840 \) watts. This conversion is crucial because it aligns with the work-energy calculations, which typically use joules and watts.

Understanding these units and conversion methods is essential in physics, particularly within real-world applications where multiple systems of measurement are used. Precise conversions ensure that energy systems operate correctly and efficiently, helping solve questions of capability such as the maximum number of passengers an elevator motor can safely accommodate.
Dynamics of Motion
Dynamics of motion explains how forces result in movement, incorporating Newton's laws of motion. An object like an elevator moving at a constant speed involves balancing forces—here, the motor's force just offsets gravitational pull, resulting in an upward vertical motion without acceleration.

The dynamics involve calculating the forces needed to achieve this balance, emphasizing how mass and force interplay over time. Given the elevator's speed and distance parameters, the motion dynamics help deduce how quickly the available power from the motor is employed. This involves using the equation \( P = \frac{W}{t} \), where \( t \) is time in seconds.

Exploring dynamics is significant for understanding constraints and operational limits, such as the example in this exercise where the maximum number of passengers is calculated. By solving for how the dynamics balance to maintain steady motion, engineers and physicists can design systems that maximize efficiency and safety while minimizing energy waste.

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Most popular questions from this chapter

A loaded grocery cart is rolling across a parking lot in a strong wind. You apply a constant force \(\vec{F}=(30 \mathrm{N}) \hat{t}-(40 \mathrm{N}) \hat{\jmath}\) to the cart as it undergoes a displacement \(\vec{s}=(-9.0 \mathrm{m}) \hat{\imath}-(3.0 \mathrm{m}) \hat{\mathrm{j}}\) How much work does the force you apply do on the grocery cart?

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A 5.00-kg package slides 1.50 \(\mathrm{m}\) down a long ramp that is inclined at \(12.0^{\circ}\) below the horizontal. The coefficient of kinetic friction between the package and the ramp is \(\mu_{\mathrm{k}}=0.310\) . Calculate (a) the work done on the package by friction; (b) the work done on the package by gravity, (c) the work done on the package by the normal force; (d) the total work done on the package. (e) If the package has a speed of 2.20 \(\mathrm{m} / \mathrm{s}\) at the top of the ramp, what is its speed after sliding 1.50 \(\mathrm{m}\) down the ramp?

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of \(1.80 \times 10^{6} \mathrm{N}\) , one \(14^{\circ}\) west of north and the other \(14^{\circ}\) east of north, as they pull the tanker 0.75 \(\mathrm{km}\) toward the north. What is the total work they do on the supertanker?

A 12 -pack of Omni-Cola (mass 4.30 \(\mathrm{kg} )\) is initially at rest on a horizontal floor. It is then pushed in a straight line for 1.20 \(\mathrm{m}\) by a trained dog that exerts a horizontal force with magnitude 36.0 \(\mathrm{N}\). Use the work-energy theorem to find the final speed of the 12 -pack if (a) there is no friction between the 12 -pack and the floor, and (b) the coefficient of kinetic friction between the 12 -pack and the floor is 0.30.

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