Acceleration of a brick, involving integrals

Click For Summary

Homework Help Overview

The problem involves calculating the net work performed on a 15 kg brick moving along an x-axis, given its acceleration as a function of position. The original poster attempts to apply integration techniques to find the work done as the brick moves from x = 0 to x = 8.0 m.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster discusses their understanding of integration and attempts to relate it to the work-energy principle. They express confusion about the integration process and seek clarification on how to approach the problem correctly.

Discussion Status

Participants are exploring different aspects of the problem, including the interpretation of the integral as the area under the curve. Some guidance has been offered regarding the calculation of the area, but there is no explicit consensus on the correct approach or answer.

Contextual Notes

The original poster mentions having difficulty with calculus concepts and has sought help from teachers without success. There is also a reference to a figure that is crucial for understanding the problem, which was initially missing.

majormaaz
Messages
61
Reaction score
1

Homework Statement


A 15 kg brick moves along an x axis. Its acceleration as a function of its position is shown in Fig. 7-32. 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?

Homework Equations


W = FΔx = max

The Attempt at a Solution


I haven't taken calculus prior to the class(am taking it now), and I was just introduced to integrals today by a friend. He showed me how to integrate another Work problem, where F = -6x. I understood that it would integrate into -3x2 + C, and I found C. My question is how would I go about this one?

Would I start as ∫F dx = ∫ (15 kg)a dx
W = ∫ 15a dx
W = 15ax
W = 15ΔaΔx ... I'm guessing this is what I should do...
W = 15(24)(80) = 2400 J

But apparently that's wrong.
I would be soooooo thankful for anyone to help me with the integration process from here.

***************************************************************
And just in case anybody tells me to ask my teacher for help, I asked both my Physics C teacher AND my calc teacher, but both refused.
 
Physics news on Phys.org
There is no picture in your post.
 
You don't need to know integration techniques. It is enough to understand that the value of the (definite) integral of a function represents the area under the graph of the function.
Here is a lot more straightforward to calculate the area (as all the segments are straight lines).
 
I mean, it seems rudimentary that the net work is the area under the curve times mass, since
the area under the curve is aΔx, and multiplying by m gives you maΔx, which equals work.
And yet, that was wrong.
 
So did you get the right answer with the new figure or not? I don't understand.
 
ahh... I just viewed the answer key. It has 1200 J, while I kept getting 2400 J.
I think I just forgot to throw the 1/2 in there. Damn. A point lost.
But thanks for helping anyways!
 

Similar threads

Replies
10
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 8 ·
Replies
8
Views
14K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
17
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K