Work problem involving a variable force

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Homework Help Overview

The problem involves a CD case sliding along a floor under the influence of a variable force described by the equation Fax = 7.0x – 2.0x^2. The case starts at rest and the discussion revolves around determining positions related to maximum work, zero work, and when the case comes to rest again.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss using integration to find work done by the force and question how to determine the maximum work and when work becomes zero.

Discussion Status

Some participants have attempted to calculate the maximum position and the corresponding work value, while others are exploring general methods for finding maxima and zeros of functions. There is a mix of attempts and questions, with no clear consensus on the correctness of the approaches yet.

Contextual Notes

Participants note the importance of providing all relevant information in initial posts to facilitate better responses. There is also mention of potential confusion regarding the relationship between positions where work is zero and when the case is at rest.

Fredimension

Homework Statement


A CD case slides along a floor in the positive direction of an x-axis while an applied force Fa acts on the case. The force is directed along the x-axis and has the x component Fax = 7.0x – 2.0x^2, with x in meters and Fax in Newtons. The case starts at rest at the position x = 0, and it moves until it is again at rest. (a) At what position is the work maximum, and (b) what is that maximum value? (c) At what position has the work decreased to zero? (d) At what position is the case again at rest?

Homework Equations


W=∫Fdx
Wmax=Fdcos1

The Attempt at a Solution


I thought of using W=∫Fdx, since the function I was given includes both the displacement and the force. However, I don't know how to find the Wmax like this.
 
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I also tried using x=0 and xf as the lower and upper limits, respectively. It left me with a 7/2(Xf)^2-2/3(Xf)^3. I don't know how to find Xf this way. Or if the work done has another different value when maxed aside from Fdcos1.
 
(a) How do you usually find the maximum of a function?

(c) How do you usually find out when a function is zero?

Also a suggestion: Make sure to post all your relevant information in your first post. If you reply to your own thread it will appear to have answers and it will be less likely that people look at it.
 
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Orodruin said:
(a) How do you usually find the maximum of a function?

(c) How do you usually find out when a function is zero?

Also a suggestion: Make sure to post all your relevant information in your first post. If you reply to your own thread it will appear to have answers and it will be less likely that people look at it.

Okay, I worked out my answers:
(a) I used the formula for finding max:c-(b^2/4a). I came out with 6.125 m
(b) I used the function which I previously integrated and inserted 6.125 m on the x variables and came out with -21.88 J.
(c) and (d) For c and d, I believe that they do mean the same, so I think they have the same answers. I equate the work done to zero, and then came out with a 5.25 m
Am I doing it right? Or is there something I did wrong?

Okay, I'll keep that in mind.
 
Fredimension said:
Okay, I worked out my answers:
(a) I used the formula for finding max:c-(b^2/4a). I came out with 6.125 m
(b) I used the function which I previously integrated and inserted 6.125 m on the x variables and came out with -21.88 J.
(c) and (d) For c and d, I believe that they do mean the same, so I think they have the same answers. I equate the work done to zero, and then came out with a 5.25 m
Am I doing it right? Or is there something I did wrong?

Okay, I'll keep that in mind.
Oh, I forgot to use the integrated function for a, which also caused a problem on my b. Thanks, I finally got it right.
 

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