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Chapter 9: Problem 48

Average Rate of Work The loaded cab of an elevator has a mass of \(3.0 \times 10^{3} \mathrm{~kg}\) and moves \(210 \mathrm{~m}\) up the shaft in \(23 \mathrm{~s}\) at constant speed. At what average rate does the force from the cable do work on the cab?

Short Answer

Expert verified
The average rate of work is \( 2.684 \times 10^{5} \text{ W} \).

Step by step solution

01

Calculate the Force

The force needed to lift the elevator is equal to its weight, which is the product of mass and gravitational acceleration. Use the equation: \[ F = m \times g \] where \( m = 3.0 \times 10^{3} \text{ kg} \) and \( g = 9.8 \text{ m/s}^2 \). Substitute the values to get \[ F = 3.0 \times 10^{3} \text{ kg} \times 9.8 \text{ m/s}^2 = 2.94 \times 10^{4} \text{ N} \].
02

Determine Work Done

Work done is the product of force and displacement in the direction of the force. Use the equation: \[ W = F \times d \] where \( F = 2.94 \times 10^{4} \text{ N} \) and \( d = 210 \text{ m} \). Substitute the values to get \[ W = 2.94 \times 10^{4} \text{ N} \times 210 \text{ m} = 6.174 \times 10^{6} \text{ J} \].
03

Calculate Average Power

Power is the rate at which work is done. Use the equation: \[ P = \frac{W}{t} \] where \( W = 6.174 \times 10^{6} \text{ J} \) and \( t = 23 \text{ s} \). Substitute the values to get \[ P = \frac{6.174 \times 10^{6} \text{ J}}{23 \text{ s}} = 2.684 \times 10^{5} \text{ W} \].

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Key Concepts

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

Force Calculation
When dealing with physics problems involving elevators or any objects in motion, calculating the force is one of the first steps. Force is essentially the interaction that changes the motion of an object. According to Newton's Second Law, force is the product of the mass of an object and its acceleration. In this specific problem, we're dealing with gravitational force because the elevator is moving vertically. Gravitational force is calculated using:
\( F = m \times g \)
Here:
  • \

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Most popular questions from this chapter

Two Chunks An \(8.0 \mathrm{~kg}\) body is traveling at \(2.0 \mathrm{~m} / \mathrm{s}\) with no external force acting on it. At a certain instant an internal explosion occurs, splitting the body into two chunks of \(4.0 \mathrm{~kg}\) mass each. The explosion gives the chunks an additional \(16 \mathrm{~J}\) of kinetic energy. Neither chunk leaves the line of original motion. Determine the speed and direction of motion of each of the chunks after the explosion. Hint: Translational momentum is conserved.

Rope Tow A skier is pulled by a tow rope up a frictionless ski slope that makes an angle of \(12^{\circ}\) with the horizontal. The rope moves parallel to the slope with a constant speed of \(1.0 \mathrm{~m} / \mathrm{s}\). The force of the rope does \(900 \mathrm{~J}\) of work on the skier as the skier moves a distance of \(8.0 \mathrm{~m}\) up the incline. (a) If the rope moved with a constant speed of \(2.0 \mathrm{~m} / \mathrm{s}\), how much work would the force of the rope do on the skier as the skier moved a distance of \(8.0 \mathrm{~m}\) up the incline? At what rate is the force of the rope doing work on the skier when the rope moves with a speed of (b) \(1.0 \mathrm{~m} / \mathrm{s}\) and \((\mathrm{c})\) \(2.0 \mathrm{~m} / \mathrm{s} ?\)

Explosion at Ground Level An explosion at ground level leaves a crater with a diameter that is proportional to the energy of the explosion raised to the \(\frac{1}{3}\) power; an explosion of 1 megaton of TNT leaves a crater with a 1 km diameter. Below Lake Huron in Michigan there appears to be an ancient impact crater with a \(50 \mathrm{~km}\) diameter. What was the kinetic energy associated with that impact, in terms of (a) megatons of TNT (1 megaton yields \(4.2 \times 10^{15} \mathrm{~J}\) ) and (b) Hiroshima bomb equivalents (13 kilotons of TNT each)? (Ancient meteorite or comet impacts may have significantly altered Earth's climate and contributed to the extinction of the dinosaurs and other life-forms.)

Cold Hot Dogs Figure \(9-32\) shows a cold package of hot dogs sliding rightward across a frictionless floor through a distance \(d=\) \(20.0 \mathrm{~cm}\) while three forces are applied to it. Two of the forces are horizontal and have the magnitudes \(F_{A}=5.00 \mathrm{~N}\) and \(F_{B}=1.00 \mathrm{~N} ;\) the third force is angled down by \(\theta=-60.0^{\circ}\) and has the magnitude \(F_{C}=4.00 \mathrm{~N}\). (a) For the \(20.0 \mathrm{~cm}\) displacement, what is the net work done on the package by the three applied forces, the gravitational force on the package, and the normal force on the package? (b) If the package has a mass of \(2.0 \mathrm{~kg}\) and an initial kinetic energy of \(0 \mathrm{~J}\), what is its speed at the end of the displacement?

Block Dropped on a Spring 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. \(9-29)\). The block becomes attached to the spring and compresses the spring \(12 \mathrm{~cm}\) before turning around. 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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