How Do You Calculate Work Done by a Changing Force?

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Homework Help Overview

The discussion revolves around calculating the work done by a changing force, specifically focusing on the area under a curve in a graph representing force versus position. Participants are exploring how to accurately compute work done when the force varies, particularly in cases involving both positive and negative work.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of calculating work as the area between the x-axis and the curve, questioning how to account for negative work in certain regions. There are attempts to apply geometric formulas for areas of triangles and trapezoids to find total work done.

Discussion Status

The conversation is active, with participants sharing their calculations and questioning their results. Some guidance has been provided regarding the use of specific area formulas and the importance of considering the signs of the areas based on their position relative to the x-axis. There is no explicit consensus on the correct total work value yet.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the resources available for reference. The original poster expresses frustration with discrepancies between their calculations and the model answer provided in their textbook.

KiNGGeexD
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I am stuck with the above question, my answer to a is 40J but the model answer is 30J
Really frustrating as the book I am ready only goes over straight line graphs
 
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Remember that the work done for this type of problem is the area between the ##x##-axis and the given curve. You got that value via adding up the areas between ##x = -2## and ##x = 4##. The first part asks you to determine the total work done. You also need to consider the "negative" work done between ##x = -4## and ##x = -2##. It also takes in account for this problem.

Try again to see if you can get ##30## J instead of ##40## J.
 
So for the first curve can I use 1/2B*h?

Because I done that for the two slopes and b*h for the square part?
 
KiNGGeexD said:
So for the first curve can I use 1/2B*h?

Because I done that for the two slopes and b*h for the square part?

Yes, you need to use the area of the triangle formula for the region under the ##x##-axis to obtain the "negative" work done.

Since we want to compute the total work done between ##x = -4## and ##x = 4##, we combine the works altogether to obtain the result. Don't forget this part.
 
So I get 4 work values?
 
So from left to right I get

20J
-10J
20J
10J

So that gives me 40J?
 
KiNGGeexD said:
So from left to right I get

20J
-10J
20J
10J

So that gives me 40J?

I didn't get those values. Let's try again.

Remember that for the region under the ##x##-axis, regardless of the values of ##x##, you get the negative work done, using the area of the triangle:

##A = -\dfrac{1}{2}|b||h|##

where the brackets indicate the absolute value of any number.

Then, for the region above the ##x##-axis, (also regardless of the values of ##x##) you use the area of the trapezoid, which states

##A = \dfrac{1}{2}|h||b_1 + b_2|##

where ##b_1## and ##b_2## are arbitrary bases of the trapezoid and ##h## is the height of the trapezoid.

You can also compute the area of the bigger region by triangle and square area formulas as you attempted. That is: determine the area of two triangles and the area of the square.

So in summary, you should get the following equation (ignoring the signs of ##x## values)

##\text{Total work done} = \text{Area between x = -4 and x = -2} + \text{Area between x = -2 and x = 4}##

where ##\text{Area between x = -4 and x = -2}## is negative while other area is positive.

Notes: Be careful about how you compute the first value. The shape is a triangle, not a square! Also be very careful about the signs of the first two values. If a region occurs under ##x##-axis, then the area is negative; otherwise, it's positive.
 
Last edited:
Ah ok I think I got it:)

Thanks for your help
 

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