# Homework Help: 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.