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Chapter 6: Problem 16

An elevator of mass \(m\) rises a vertical distance \(h\) with upward acceleration equal to one-tenth \(g .\) Find an expression for the work the elevator cable does on the elevator.

Short Answer

Expert verified
The work done by the elevator cable on the elevator is \(1.1mgh\)

Step by step solution

01

Determine the net force on the elevator

The net force on the elevator is the sum of the gravitational force acting on the elevator and the additional force needed for the elevator to accelerate upward. This can be expressed as \(F = m(g + a)\), where \(m\) is the mass of the elevator, \(g\) is the acceleration due to gravity, and \(a\) is the upward acceleration of the elevator. Given that the upward acceleration \(a\) is one-tenth of \(g\), the net force can be rewritten as \(F = m(g + 0.1g) = 1.1mg\).
02

Calculate the work done by the cable on the elevator

The work done on an object by a force is equal to the force times the distance the object moves in the direction of the force. Here the distance moved by the elevator is \(h\) and the force is in the upward direction, thus the work done can be given by \(W = Fh = 1.1mgh\). Where \(W\) is the work done, \(F\) is the force, \(m\) is the mass, \(g\) is the acceleration due to gravity and \(h\) is the height the elevator moved.

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