This problem is driving me crazy. A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.15 and the push imparts an initial speed of 3.5m/s?
The answer is 4.2m but I don't know how to get it.

Question

This problem is driving me crazy.  A box is given a push so that it slides across the floor. How far will it go, given that the coefficient of kinetic friction is 0.15 and the push imparts an initial speed of 3.5m/s?
The answer is 4.2m but I don't know how to get it.

anonymous · Answer

d=v^2/2ug
d=(3.5)^2/(0.15)(9.8m/s^2)

anonymous · Answer

this should give approx. 4.2m

anonymous · Answer

How did you get $$x -x_{0} = \frac{v^{2}}{2\mu g}$$?

anonymous · Answer

Oh wait, I see how. Thanks!

anonymous · Answer

ok so
 F=umg work done by frictional force is W=umgd
andwe know that KE=1/2mv^2
1/2mv^2=umgd

anonymous · Answer

:)