What equations would I use to solve for work and power delivered?

1. Oct 4, 2011

george256

A 6 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 6 + 4x - 3x2, where Fx is in newtons and x is in meters.

(a) Find the work done by the force as the particle moves from x = 0 to x = 2 m.
in J
(b) Find the power delivered to the particle when it is at x = 2 m.
in W

2. Oct 4, 2011

Pengwuino

How do you determine the work done on an object given a certain force acting over a certain distance? This should be a very basic equation that you should know the equation for.

Also, the units seem to be a bit odd, are you sure that's the correct form of the force?

3. Oct 4, 2011

george256

fx is 6+4x-3x^2 sorry about that. and would i use the equation w= integral of fx from distance initial to distance final?

4. Oct 4, 2011

Pengwuino

Yes. There you go. Your integration bounds are given the initial and final distances.

5. Oct 4, 2011

george256

ok now do i have to integrate the fx or is that already accomplished and i should just plug in the numbers, because when i integrate that i get -x^3+2x^2+6x and when i integrate from 0 to 2 i get the answer incorrect.

6. Oct 4, 2011

Pengwuino

No, you must integrate the function. Check your work, you integrated correctly.

7. Oct 4, 2011

george256

and also how can i find power without a time given

8. Oct 4, 2011

george256

For that after i integrate i get 20 is that incorrect?

9. Oct 4, 2011

Pengwuino

No, that's not correct.

I didn't even notice the part asking about power. I really think you haven't written down the problem correctly. Are you sure you don't mean $x(t) = 6 + 4t - 3t^2$?

10. Oct 4, 2011

george256

that is the exact copy of my problem. i just looked it over and i am not sure where i am going wrong when i integrate all i should have to do is plug 2 in as x correct because 0 will just yield 0 when plugged in as x

11. Oct 4, 2011

Pengwuino

You don't have enough information to determine the power.

When you integrate, you should get the work done = 12J, not 20J (check your algebra). I highly doubt you have copied the question correctly. There is no way to determine the power with what you are given.

12. Oct 4, 2011

george256

that is what i thought and i just double checked that. when you integrate is it products of time inital minus time final? is that where i am making my mistake. I have not integrated in a very long time.

13. Oct 4, 2011

george256

never mind i realize what i did wrong in integration. stupid mistake. I wish i could figure out the power though

14. Oct 4, 2011

george256

thanks for your help. i appreciate it. i just gave up on the power. if you could not figure it out i knew it must not be possible.

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