# Question regarding work.

1. Feb 22, 2007

### ssb

My TA gave us a conceptual example today that I dont 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

2. Feb 22, 2007

### arildno

No, it is not.
The net work done equals the change in the object's kinetic energy, which certainly is non-zero.

3. Feb 22, 2007

### ssb

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?

4. Feb 22, 2007

### Staff: Mentor

You have the acceleration as a function of time. What would you need to do to get the velocity as a function of time?

5. Feb 22, 2007

### ssb

take the derivative right? so it would be 6x right? sorry im new to this stuff

6. Feb 22, 2007

### ssb

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 im thinking i did something wrong.

7. Feb 22, 2007

### Staff: Mentor

Last edited by a moderator: Apr 22, 2017
8. Feb 22, 2007

### ssb

ahhhh cool it is starting to make sence. Thanks!

9. Feb 22, 2007

### turdferguson

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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