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Chapter 7: Problem 75

An elevator cab has a mass of \(4500 \mathrm{~kg}\) and can carry a maximum load of \(1800 \mathrm{~kg}\). If the cab is moving upward at full load at \(3.80 \mathrm{~m} / \mathrm{s}\), what power is required of the force moving the cab to maintain that speed?

Short Answer

Expert verified
The required power is 234.47 kW.

Step by step solution

01

Calculate Total Mass

First, determine the total mass of the elevator cab when fully loaded. This is the sum of the elevator cab mass and the maximum load:\[\text{Total mass} = 4500\, \text{kg} + 1800\, \text{kg} = 6300\, \text{kg}\]
02

Identify Speed and Gravity

We are given that the cab moves upward at a speed of \(3.80\, \mathrm{m/s}\), and we need the gravitational acceleration \(g\), which is \(9.81\, \mathrm{m/s}^2\). We'll use these in the later steps.
03

Determine Force Required

To maintain the upward speed, the force needed to overcome gravity is equal to the weight of the fully loaded elevator cab. Calculate the force (weight) as follows:\[F = \text{Total mass} \times g = 6300\, \text{kg} \times 9.81\, \text{m/s}^2 = 61703\, \text{N}\]
04

Calculate Power

Power is the rate of doing work, which can be calculated using the formula:\[P = F \times v\]where \(v\) is the velocity. Substitute the known values:\[P = 61703\, \text{N} \times 3.80\, \text{m/s} = 234470.4\, \text{W}\]
05

Convert Power to Kilowatts

Power is commonly expressed in kilowatts. To convert watts to kilowatts, divide by 1000:\[P = \frac{234470.4\, \text{W}}{1000} = 234.47\, \text{kW}\]

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Key Concepts

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

Elevator Mechanics
Elevator systems are a fascinating combination of modern engineering designs, allowing for safe vertical transportation. In these systems, the cab, which in this case with full load includes a mass of 4500 kg for the cab plus 1800 kg of maximum load, must be moved efficiently and safely. Elevators work with a system of ropes and pulleys. The motor provides the necessary force to overcome gravity and additional mechanical oppositions, like friction.

For an elevator moving upwards, the calculations must ensure that the motor can handle the total load weight. This brings us to the force exerted by the motor, which must counteract the downward pull of gravity on the cab and its contents. Calculating this force is crucial in determining the power requirements.
Force and Motion
Force and motion are deeply interconnected. In the case of our elevator, this relationship is fundamental. Force is required to move the elevator upwards, and this force must be greater than the gravitational force pulling the cab downwards.

Given the total mass of the elevator and the person or goods inside, calculated to be 6300 kg, we use the gravitational constant (9.81 m/s虏) to find the force due to gravity, which is 61703 N for our scenario. This force ensures the motion of the elevator is smooth and uninterrupted. Essentially, for seamless motion, the engine must provide an equivalent force to balance this gravitational pull when moving at constant speed.
Kinetic Energy
Kinetic energy relates to the energy an object possesses due to its motion. For an elevator moving at a constant speed of 3.80 m/s, the energy is calculated using the formula: \[ KE = \frac{1}{2}mv^2 \]where \(m\) is the total mass, and \(v\) is the velocity. While the kinetic energy itself does not change when speed is constant, it's important for understanding how much power is required initially to reach and maintain this motion. When calculating power, as we did with \[ P = F \times v \]we can see that force in motion aligns with the changing kinetic energy as the elevator reaches the set speed. In practical terms, the power ensures the elevator's motor can efficiently handle the energy transitions without causing operational delays or safety hazards in usage.

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Most popular questions from this chapter

A luge and its rider, with a total mass of \(85 \mathrm{~kg}\), emerge from a downhill track onto a horizontal straight track with an initial speed of \(37 \mathrm{~m} / \mathrm{s}\). If a force slows them to a stop at a constant rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2},(\mathrm{a})\) what magnitude \(F\) is required for the force, (b) what distance \(d\) do they travel while slowing, and (c) what work \(W\) is done on them by the force? What are (d) \(F,(\mathrm{e}) d\), and \((\mathrm{f}) W\) if they, instead, slow at \(4.0 \mathrm{~m} / \mathrm{s}^{2} ?\)

The force on a particle is directed along an \(x\) axis and given by \(F=F_{0}\left(x / x_{0}-1\right)\). Find the work done by the force in moving the particle from \(x=0\) to \(x=2 x_{0}\) by (a) plotting \(F(x)\) and measuring the work from the graph and (b) integrating \(F(x)\).

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?

The only force acting on a \(2.0 \mathrm{~kg}\) body as it moves along a positive \(x\) axis has an \(x\) component \(F_{x}=-6 x \mathrm{~N}\) with \(x\) in meters. The velocity at \(x=3.0 \mathrm{~m}\) is \(8.0 \mathrm{~m} / \mathrm{s}\). (a) What is the velocity of the body at \(x=4.0 \mathrm{~m} ?\) (b) At what positive value of \(x\) will the body have a velocity of \(5.0 \mathrm{~m} / \mathrm{s} ?\)

A \(250 \mathrm{~g}\) block is dropped onto a relaxed vertical spring that has a spring constant of \(k=2.5\) \(\mathrm{N} / \mathrm{cm}\) (Fig. 7-44). The block becomes attached to the spring and compresses the spring \(12 \mathrm{~cm}\) before momentarily stopping. While the spring is being compressed, what work is done on the block by (a) the gravitational force on it and (b) the spring force? (c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) (d) If the speed at impact is doubled, what is the maximum compression of the spring?
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