# Homework Help: Force applied to an object

1. Feb 12, 2014

### s.dyseman

1. The problem statement, all variables and given/known data

A net force along the x-axis that has x-component Fx=−12.0N+(0.300N/m2)x2 is applied to a 3.20kg object that is initially at the origin and moving in the -x-direction with a speed of 8.60m/s

What is the speed of the object when it reaches the point x = 9.00m ?

2. Relevant equations

F=ma
.5mv^2=Ke

3. The attempt at a solution

Not really sure where to go with this. I can find, using the basic N2 formula, that the acceleration of the object changes according to the Fx equation given divided by mass, giving an acceleration of 3.84 m/s at x=9.00m.

However, I'm not sure how to find how far the object travels in the -x direction. Is the acceleration I have calculated at x=9 instantaneous or average? I'm not sure how to apply the kinetic energy to this problem either. Initial energy is 118.34 J, however, as the force applied is changing, isn't the energy going to continuously change as well? We can't apply any sort of energy conservation to this if that is the case.

I can understand that I need to find how far the object travels in the -x direction, then calculate the change in velocity over the distance it travels back toward the x=9 mark... I'm just lost as to how to get there...

2. Feb 12, 2014

### BvU

1. You need more equations. I read motion under influence of a force that is dependent on position, so not constant. Since you want speed and position in there and F is not constant, you want very basic stuff: definitions of v and a. Everything is instantaneous. Averages come in only when some things are constant or when explicitly asked for.

2. At t=0 the thing moves to the left and even accelerates in that direction. Then the x2 term kicks in and hopefully that's enough to get on the positive side of the x axis. Who will tell... Well: your worked out equation(s). About the need to split it up: let's get going first and decide later.

3. Feb 12, 2014

### Staff: Mentor

Are you studying work-energy relationships in your course yet? The force is an even function of x, so how much work is done by the force in moving the mass from x = 0 to x = 9 m (even if the mass moves to the left first)? How is this related to the increase in kinetic energy of the mass?

Chet

4. Feb 12, 2014

### s.dyseman

So, if I understand, I can simply integrate the force equation given from 0 to 9 to find the total work done. From what I gather, the change in total energy must equal the change in kinetic energy from 0 to 9, so we can just set the total work = (1/2)mv^2 to find the velocity at x=9?

5. Feb 12, 2014

### s.dyseman

Also, due to the initial velocity, do we not need to find where the velocity of the object is equal to zero (which would be in the -x direction), so that we can integrate from that -x value to x=9?

6. Feb 12, 2014

### BvU

should be: total work = change in (1/2)mv^2

(Good thing Chet jumped in, I would have gotten us stuck very quickly!)

7. Feb 12, 2014

### s.dyseman

So just to clarify, the delta would not be from 0 to 9, but from the -x value where v=0 to 9, right?

And yes, a big thanks to both you and Chet.

8. Feb 12, 2014

### BvU

The beauty of this energy thing is that it doesn't matter. Valid in both cases. The easier one is from 0 to 9, because you don't know the turning point (you can always check afterwards...;-)

9. Feb 12, 2014

### s.dyseman

Should I be getting a negative value after integrating from 0 to 9? I'm getting -35.1. This would lead to a negative velocity at x=9, correct? Is that possible?

10. Feb 12, 2014

### s.dyseman

Or should I be subtracting that value from the initial KE?

11. Feb 12, 2014

### s.dyseman

Ah, I got it! Subtract that value from KE-initial and set it equal to the KE equation. V=7.21. Thank you guys so much!

12. Feb 12, 2014

### BvU

Yep. Got same speed. Thanks to Chet.