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Chapter 7: Problem 75

What is the power of the force required to move a 4500 \(\mathrm{kg}\) elevator cab with a load of \(1800 \mathrm{~kg}\) upward at constant speed \(3.80 \mathrm{~m} / \mathrm{s} ?\)

Short Answer

Expert verified
The power required is approximately 234.47 kW.

Step by step solution

01

Determine Total Mass

First, calculate the total mass of the elevator cab and its load. The mass of the elevator cab is 4500 kg, and the load is 1800 kg. Thus, the total mass \( m \) is given by \( m = 4500 \, \text{kg} + 1800 \, \text{kg} = 6300 \, \text{kg} \).
02

Identify the Gravitational Force

The gravitational force \( F_g \) acting on the total mass is calculated using the equation \( F_g = m \cdot g \), where \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity. Plugging in the total mass, \( F_g = 6300 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 61703 \, \text{N} \).
03

Calculate Power Required

Power \( P \) is defined as the work done per unit time. When moving at constant speed, power can be calculated by \( P = F_g \cdot v \), where \( v = 3.80 \, \text{m/s} \) is the constant speed of the elevator. Thus, \( P = 61703 \, \text{N} \times 3.80 \, \text{m/s} = 234,472.4 \, \text{W} \).
04

Convert to Kilowatts

Since power is often expressed in kilowatts, convert the power from watts to kilowatts by dividing by 1000. Therefore, \( P = \frac{234,472.4 \, \text{W}}{1000} = 234.4724 \, \text{kW} \).

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Key Concepts

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

Gravitational Force
Gravitational force is a fundamental concept that is key to understanding many physical processes. It is the force by which a planet or other celestial body draws objects toward its center. This force gives weight to physical objects and causes them to fall if dropped. In the context of the elevator cab exercise, the gravitational force is the force acting downward on both the cab and its load.
To calculate this force, we use the formula:
  • Gravitational Force (\( F_g \)) = mass (\( m \)) x acceleration due to gravity (\( g \))
The acceleration due to gravity is a constant, approximately 9.81 m/s虏 on the surface of the Earth.
By multiplying the total mass (cab + load) by this constant, we find the gravitational force exerted on the elevator cab. This force is what we need to counter to lift the elevator upwards.
Work and Energy
The concepts of work and energy are crucial in physics as they describe how objects move and interact with forces. Work is done when a force causes an object to move in the direction of the force. Mathematically, work done is the product of the force applied and the distance moved in the direction of the force.
In the case of an elevator, when it moves upwards against the gravitational pull, work is performed by the lifting mechanism.
Energy, on the other hand, is the capacity to do work. Power, a related concept, is the rate at which work is done or energy is transferred.
  • Power (\( P \)) = Work done per unit time
In our problem, to maintain a constant speed, the elevator must perform enough work to overcome gravitational force without accelerating. Calculating the required power helps us understand the energy expenditure needed to maintain this constant upward motion.
Constant Speed Motion
Moving at a constant speed means that the velocity of the object does not change with time. In physics, this implies that the net force acting on the object is zero. Thus, all forces are balanced, leading to no acceleration.
For the elevator, moving upward with a constant speed of 3.80 \( ext{m/s} \) indicates that the force supplied by the elevator mechanisms is exactly equal to the gravitational force acting downward.
  • This means that all energy supplied to the system goes into counteracting gravity.
  • There is no net increase in kinetic energy since speed remains constant.
Understanding constant speed motion is crucial for calculating the power needed. Since there is no change in speed, the power required is simply the product of gravitational force and speed, with no allowances for accelerating the elevator.

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Most popular questions from this chapter

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 \(100 \mathrm{~kg}\) block is pulled at a constant speed of \(5.0 \mathrm{~m} / \mathrm{s}\) across a horizontal floor by an applied force of \(122 \mathrm{~N}\) directed \(37^{\circ}\) above the horizontal. What is the rate at which the force does work on the block?

A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of \(10.0\) m: (a) the initially stationary spelunker is accelerated to a speed of \(5.00 \mathrm{~m} / \mathrm{s} ;(\mathrm{b})\) he is then lifted at the constant speed of \(5.00 \mathrm{~m} / \mathrm{s} ;\) (c) finally he is decelerated to zero speed. How much work is done on the \(80.0 \mathrm{~kg}\) rescuee by the force lifting him during each stage?

How much work is done by a force \(\vec{F}=(2 x \mathrm{~N}) \hat{\mathrm{i}}+(3 \mathrm{~N}) \hat{\mathrm{j}}\), with \(x\) in meters, that moves a particle from a position \(\vec{r}_{i}=\) \((2 \mathrm{~m}) \hat{\mathrm{i}}+(3 \mathrm{~m}) \hat{\mathrm{j}}\) to a position \(\vec{r}_{f}=-(4 \mathrm{~m}) \hat{\mathrm{i}}-(3 \mathrm{~m}) \hat{\mathrm{j}}\) ?

A breadbox is made to move along an \(x\) axis from \(x=0.15 \mathrm{~m}\) to \(x=1.20 \mathrm{~m}\) by a force with a magnitude given by \(F=\exp \left(-2 x^{2}\right)\), with \(x\) in meters and \(F\) in newtons. (Here exp is the exponential function.) How much work is done on the breadbox by the force?
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