Calculating Rail Gun Distance for Escape Velocity

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The discussion revolves around calculating the distance a conducting bar must travel along rails to achieve Earth's escape velocity of 11.2 km/s, given a magnetic force of 588 N acting on a 25 kg bar. The initial attempt at the solution incorrectly applied kinematic equations, leading to an excessively large distance calculation of over 5 million meters. Participants suggest using the correct kinematic equation, V^2 = 2a*x, which simplifies the problem by eliminating the need to calculate time. The conversation emphasizes the importance of correctly applying physics formulas to derive the accurate distance required for escape velocity. Ultimately, the correct approach focuses on the relationship between acceleration, distance, and final velocity.
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Homework Statement



A conducting bar with mass=25.0 kg and length L=51.0 cm slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current I= 2400A in the rails and bar, and a constant, uniform, vertical magnetic field B = 0.480 T fills the region between the rails.
See figure:
railgun.jpg


It has been suggested that rail guns based on this principle could accelerate payloads into Earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the Earth (11.2 km/s).
For simplicity, assume the net force on the object is equal to the magnetic force, as in parts A and B, even though gravity plays an important role in an actual launch into space.


Homework Equations



I know the magnetic force is 588N.
F=ma
V = at
X = at^2


The Attempt at a Solution



588/25 = 23.52 m/s^2
11200 m/s = 23.52 m/s^2 * t
t = 476.190

x = 23.52 m/s^2 * 476.19^2
5333322.666672 m

Am I missing something, it says that is incorrect...
 
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I think the best approach here might be with the kinematics eqn that relates velocity to distance along which body is uniformly accelerated, since Vi=0 (body at rest): V^2=2a*x
 
ah, my formulas were wrong... it's x=1/2at^2 and v=at... I should've caught that...
 
you miss my point maybe. there is no need to mess with time, but certainly you can get the right answer using that approach. Try it both ways!
 
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