Calculating Work Done by Varying Force on a Moving Block - Easy Physics Problem

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In summary, the conversation is about a physics problem involving a 3.0 kg block moving on a frictionless surface. The question is how much work is done by a force that varies with position as the block moves from the origin to x = 8.0 m. The person asking the question initially thought the solution was to find the area under the graph, but was getting incorrect answers. After some discussion and clarification, it is determined that the correct approach is to take the areas around the F=0 line, resulting in a work done of 25 units. The person asking the question apologizes for their misunderstanding and thanks the expert for their help.
  • #1
DizBelieF14
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I honestly feel like a complete idiot asking this, since I know it's really easy, but I just can't get the answer. The question is...

A 3.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown in Figure 7-38 (attached). How much work is done by the force as the block moves from the origin to x = 8.0 m?

I thought this problem was as simple as finding the area under the curve, because it's force versus distance. Doing this, I got -45 because the bottom of the graph goes into the negatives (or 105 if negatives don't mean anything). Neither of these answers are correct (using WebAssign for homework), and I can't figure out why. Am I just being really stupid with the math? Does the mass even matter, I thought it didn't because you're just integrating? Sorry for the stupid question, I just don't have anyone else to ask... :shy:
 

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  • #2
I can't see the attachment, but basically, the work should be [tex]W = \int_{0}^8 F(x)dx[/tex], where F(x) is your graph. The integral actually equals the change in potential, U(8) - U(0), where the potential U(x) is the antiderivative of F(x).
 
  • #3
Can you give us an expression for F - given that we can't see the attachment yet?

You must have one, to be able to integrate? Or is your graph just straight lines?
 
  • #4
It is simply a graph with straight lines. I thought that you simply break them up into rectangles and triangles and just add the areas..I really hope my simple math is just wrong, because otherwise I have no idea..Heres the direct link to the graph.

http://www.webassign.net/hrw/07_33.gif
 
  • #5
It must be your maths. I can't get 105 or 45 no matter what.

Remember you´re taking the areas around the F=0 line.
 
  • #6
Oh, around the 0? Okay, 25, I got it. I thought you just took it to the bottom of the graph itself. Sorry for the stupid question, I am a freshman in high school in physics, so I don't pick up on this stuff easily by myself. Thank you very much. :biggrin:
 
  • #7
Easy mistake from the way that graph is drawn!
 
  • #8
DizBelieF14 said:
Oh, around the 0? Okay, 25, I got it. I thought you just took it to the bottom of the graph itself. Sorry for the stupid question, I am a freshman in high school in physics, so I don't pick up on this stuff easily by myself. Thank you very much. :biggrin:

There are no stupid questions. :smile:
 
  • #9
But there are, at times, stupidly presented graphs... ;)
 

FAQ: Calculating Work Done by Varying Force on a Moving Block - Easy Physics Problem

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