Work Done by a General Variable Force

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SUMMARY

The discussion centers on calculating the net work performed on an 8 kg brick moving along the x-axis, with its acceleration defined as a function of position. Initially, the user attempted to find the area under the acceleration curve and multiply it by the mass, but later confirmed that the method was correct after correcting a typographical error. Additionally, the user inquired about alternative methods for solving the problem, specifically referencing the use of slope in relation to the acceleration graph.

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  • Understanding of Newton's laws of motion
  • Familiarity with the concept of work and energy in physics
  • Knowledge of graph interpretation, specifically acceleration vs. position graphs
  • Basic calculus concepts, particularly area under a curve
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  • Study the relationship between acceleration, velocity, and displacement in physics
  • Learn how to calculate work done using the area under a force vs. displacement graph
  • Explore advanced techniques for analyzing motion, such as using derivatives to find slopes
  • Investigate alternative methods for solving work-energy problems in classical mechanics
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Students of physics, educators teaching mechanics, and anyone interested in understanding the principles of work and energy in motion analysis.

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A 8 kg brick moves along an x axis. Its acceleration as a function of its position is shown in Figure 7-37. What is the net work performed on the brick by the force causing the acceleration as the brick moves from x = 0 to x = 8.0 m?

07_34.gif


What I tried to do is find the area and multiply by the mass, but it doesn't work.

Please help.


Edit: Never mind it worked. I just typed it in wrong. But I have another concern.

Is there any additional ways to solve this problem other than the one I mentioned? I was looking at someone else's work and that person did something with the slope. How does that work?
 
Last edited:
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By looking at it I could see that if they take info from the slope they would appear to be taking a longer route, finding the slope just cancels the displacements.
 

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