How can we accurately calculate work using changing acceleration?

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Homework Help Overview

The discussion revolves around calculating work done on an object with changing acceleration, specifically given an acceleration function of 3t^2 for a 3kg object over the first 2 seconds. Participants are exploring the implications of this setup and questioning the validity of the initial conclusion that the work done is zero.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to derive velocity from the acceleration function and are discussing the relationship between work, force, and distance. Questions arise about the correct method to calculate work when acceleration is not constant.

Discussion Status

There is an ongoing exploration of the concepts involved, with some participants suggesting the need to integrate to find velocity and others expressing uncertainty about their calculations. Guidance has been offered regarding the change in kinetic energy as a more appropriate approach to determining work.

Contextual Notes

Participants are grappling with the implications of changing acceleration on work calculations and are considering the assumptions made in the original problem setup. There is a noted lack of consensus on the correct interpretation of the work done in this scenario.

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My TA gave us a conceptual example today that I don't understand. She was explaining how to derive work if you are given acceleration.

She wrote on the board that an object accelerates with 3t^2. the object weighs 3kg, starts at t=0. She asked us how much work is done on the object durring the first 2 seconds and then she said the answer was zero.

Is this right? what am I not getting:confused: :confused:
 
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No, it is not.
The net work done equals the change in the object's kinetic energy, which certainly is non-zero.
 
arildno said:
No, it is not.
The net work done equals the change in the object's kinetic energy, which certainly is non-zero.

Thanks I knew she was crazy. That makes me feel better.

Do you know how to calculate this in joules by the way so I can go back to her and talk to her about this?
 
You have the acceleration as a function of time. What would you need to do to get the velocity as a function of time?
 
take the derivative right? so it would be 6x right? sorry I am new to this stuff
 
i tried solving for the position function which is just 6. graphicly speaking, a graph of y = 6 from 0 to 2 would cover 12 meters.
since work is force x distance and force is mass x acceleration:

work = (3kg * 3x^2) * (12)
so at time = 2, the work would be 432 joules?

Im sorry but I am thinking i did something wrong.
 
Last edited by a moderator:
ahhhh cool it is starting to make sence. Thanks!
 
ssb said:
i tried solving for the position function which is just 6. graphicly speaking, a graph of y = 6 from 0 to 2 would cover 12 meters.
since work is force x distance and force is mass x acceleration:

work = (3kg * 3x^2) * (12)
so at time = 2, the work would be 432 joules?

Im sorry but I am thinking i did something wrong.

That would work if your force was constant, but it acceleration is changing and so is force. The best way is to figure out the change in kinetic energy
 

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