Work problem involving a variable force

In summary, the maximum work is achieved at 6.125 m with a value of -21.88 J. The work decreases to zero at 5.25 m, which is also where the case is at rest again.
  • #1
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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  • #2
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.
 
  • #3
(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.
 
  • Like
Likes Fredimension
  • #4
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.
 
  • #5
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.
 

Related to Work problem involving a variable force

What is a variable force?

A variable force is a force that changes in magnitude or direction as it acts on an object. It can be represented by a mathematical function, and is typically dependent on one or more variables.

What types of problems involve variable forces?

Problems involving variable forces can include situations where the force changes over time, changes depending on the position of the object, or changes due to external factors such as friction or air resistance.

How do you calculate work involving a variable force?

The work done by a variable force can be calculated by integrating the force function with respect to the variable it depends on. This will give the total work done over a given distance or time interval.

What is the equation for work involving a variable force?

The equation for work involving a variable force is W = ∫F(x) dx, where F(x) is the force function and dx is the change in the variable over which the force is acting.

How can I graph a variable force function?

To graph a variable force function, plot the values of the force on the y-axis and the corresponding values of the variable on the x-axis. This will create a curve that represents the force as it changes with respect to the variable.

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