Velocity of a steel ball bearing due to magnetic force

[jonc1258]

Homework Statement

Find the final velocity of a 5g steel ball bearing being pulled by a neodymium magnet with a pull force of 6.1 pounds (27.1 Newtons) in a straight line on a frictionless surface after being pulled for .1 meters

Homework Equations

F=MA

The Attempt at a Solution

f=ma
27.1 = .05a
a = 542m/s^2
The bearing travels for .1 meters, which is 1.85E-4 as far as it will travel in the first second (if the above solution for acceleration is correct), so t must be 1.85E-4 seconds.
v = (.1 m)/(1.85E-4 s) = 542m/s

As you can probably see by how unrealistic the numbers I came up with are, I'm not sure what to do. Could you maybe think about this in terms of kinetic energy? I got to feeling that this problem is more complicated than I initially thought.

 

[haruspex]

27.1 = .05a

How many g in a kg?

The bearing travels for .1 meters, which is 1.85E-4 as far as it will travel in the first second (if the above solution for acceleration is correct), so t must be 1.85E-4 seconds.

No, it doesn't work like that. What equation do you know relating distance, speed and uniform acceleration?

 

[jonc1258]

27.1 = .005a
a = 5420m/s^2

v$^{2}_{f}$ = v$^{2}_{i}$ + 2ad
v$^{2}_{f}$ = (0) + 2(5420)(.1)
vf^2 = $\sqrt{1084}$
v=32.9m/s

Yes?

 

[jonc1258]

What if the first bearing were the trigger for a Gauss rifle?
http://sci-toys.com/scitoys/scitoys/magnets/gauss_rifle/ready_to_fire.jpg
If the second magnet was .1 meters in front of the first magnet and there was another bearing lying against the second magnet (like in the picture), wouldn't the momentum of the trigger transfer to the second bearing? And the second bearing would accelerate because of the force of the 3rd magnet in the setup, but wouldn't it be pulled back by the attraction of the second magnet?

 

[haruspex]

27.1 = .005a
a = 5420m/s^2

v$^{2}_{f}$ = v$^{2}_{i}$ + 2ad
v$^{2}_{f}$ = (0) + 2(5420)(.1)
vf^2 = $\sqrt{1084}$
v=32.9m/s

Yes?

Yes.