Acceleration of a brick, involving integrals

AI Thread Summary
The discussion revolves around calculating the net work performed on a 15 kg brick as it moves along the x-axis, given its acceleration as a function of position. The user initially attempts to apply integration to find the work done but struggles with the process and ends up with an incorrect answer of 2400 J. After reviewing the correct figure and realizing the need to include a factor of 1/2 in the calculation, the user acknowledges that the correct answer is 1200 J. The conversation highlights the importance of understanding the relationship between the area under the acceleration curve and work done. Overall, the integration process and the significance of accurately interpreting the graph are emphasized in solving the problem.
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!
 
Back
Top