Acceleration of a brick, involving integrals

In summary, the net work performed on a 15 kg brick moving along an x axis from x = 0 to x = 8.0 m by a force causing the acceleration represented in Fig. 7-34 is 1200 J. This can be calculated by finding the area under the curve (aΔx) and multiplying it by the mass, taking into account the 1/2 factor.
  • #1
majormaaz
62
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.

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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.
 
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  • #2
There is no picture in your post.
 
  • #4
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).
 
  • #5
  • #6
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.
 
  • #7
So did you get the right answer with the new figure or not? I don't understand.
 
  • #8
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!
 

FAQ: Acceleration of a brick, involving integrals

What is the formula for acceleration of a brick involving integrals?

The formula for acceleration of a brick involving integrals is a = ∫(F/m)dt, where a is the acceleration, F is the force applied on the brick, m is the mass of the brick, and ∫ represents the integral symbol.

How is the acceleration of a brick calculated using integrals?

The acceleration of a brick can be calculated using integrals by integrating the force applied on the brick over time and dividing it by the mass of the brick.

What is the significance of using integrals to calculate acceleration of a brick?

Using integrals to calculate acceleration of a brick allows for the consideration of changing forces and the effect of time on the acceleration of the brick, providing a more accurate measurement.

Can integrals be used to calculate the acceleration of a brick in different scenarios?

Yes, integrals can be used to calculate the acceleration of a brick in different scenarios as long as the force and mass of the brick are known and the integral can be evaluated.

How can the acceleration of a brick be graphically represented using integrals?

The acceleration of a brick can be graphically represented using integrals by plotting the force applied on the brick over time and finding the area under the curve, which represents the change in velocity and therefore the acceleration of the brick.

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