Kinetic Friction Question?

Question

A block of mass m moving with a initial velocity V moves over a rough surface with a coeffcient of kinetic fricton u (mew). How long does it take for the block to come to a complete stop

A. V^2/2ug
B. V^2/2mug
C. V^2/ug
D. V/2ug
E. V/2mug

Answer 1

The work done by friction will be equal to the starting kinetic energy
m*V^2/2=m*g*u*d
solve for d
V^2/(2*g*u)=d
the distance is equal to the average V multiplied by time
or
d=V*t/2
plug into above
V^2/(2*g*u)=V*t/2
solve for t
V/(g*u)=t
t=V/g*u

unless I made an algebra error, the answer doesn't appear.
The only choice that has units of time is D.

j