3813.75 Joules of work is being done on a 150 lb object while it is moved 75 ft across a horizontal plane, what is the final velocity if the coefficient of kinetic friction is 0.25 and the initial velocity is 1 ft / s?

Answer 1

vf^2=vi^2+2ad

but a=force/mass=mu*mg/m=mu*g
solve for vf

change lb to kg, 75 f to meters. vi to m/s

Answer 2

Thanks for the help, but I still need help on more one piece.

How do I get the final velocity or get the ^2 out of the equation. Here are my following numbers:
Vi= 0.3048 m/s
Vf= ?
d = 22.86 m
a = 166 m/s^2
t = Not required

Based on the numbers I have would I set it up this way???

Vf^2= 0.3048^2 + 2*(166*22.86)

This process comes out to a impossible number... Please help!