w4_elevator - W done by the force F ( x ) in moving a...

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1 Problem. Going up? When a body moves a distance d along a straight line as a result of being acted on by a force of constant magnitude F in the direction of motion, then the work W done by the force is W = Fd . We will generalize the definition of work to the case of variable force. Given a force function F ( x ) defined and continuous at each point x of the straight line segment [ a, b ] , we will define the work W done by this variable force in moving a particle along the x -axis from the point x = a to the point x = b . Partition the interval [ a, b ] into n subintervals with the same length Δ x . Choose an arbitrary point c k in the k ’th interval [ x k - 1 , x k ] . Approximate the work Δ W k done by the force from the point x = x k - 1 to the point x = x k by Δ W k F ( c k x. We approximate the total work by summing from 1 to n , so W n X k =1 F ( c k x. This is a Riemann sum for F ( x ) . When Δ x 0 , the sum approaches the definite integral of F ( x ) from a to b . Therefore, we are motivated to define the work
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Unformatted text preview: W done by the force F ( x ) in moving a particle from a to b to be Z b a F ( x ) dx . Problem. An electric elevator with a motor at the top has a multistrand cable weighing 2 kg/m. When the car is at the rst oor, then 60 meters of cable are paid out. When the car is at the top oor, then effectively meters are paid out. a) What is the force exerted by the motor to just lift the cable? Hint: Newtons second law says that the force is equal to mass times acceleration, F = ma , with units measured in Newtons [N=kg m/s 2 ]. You may take the free gravitational accelera-tion to be g = 10m/s 2 . b) How much work does the motor do just lifting the cable when it takes the car from the rst oor to the top?...
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This note was uploaded on 10/31/2011 for the course MATH 1910 at Cornell University (Engineering School).

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