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A car with a mass of 100 kg, with initial v = 1 m/s, has 60 Joules of energy added for 50 m. What is the final velocty?

I'm using KE = 1/2 mv^2 to find initial KE. Then add 600 J to this for the final KE. Thus, W = KFf - KEi;

KEf = W + KEi

and KEf = 1/2mv^2

v^2 = (2(KEf))/m

3 answers

Method looks fine to me but is it 60 or 600 Joules that you add?
The "for 50 m" part of your question confuses me. The distance over which the energy is added makes no difference in the final velocity.

You method is correct, but as Damon noted, you have two different numbers for the added energy.
Oop's, my bust. It's 60 Joules. I believe the 50m was thrown is a curve to throw me off, as as the '60J' = W.

Otherwise it would have been say '60N over 50m = 3000J'.

Thanks guys.