Question

A 45kg sled is coasting with constant velocity at 3.0m/s over a patch of smooth and level ice. It enters a very rough section of ice that is 11m long in which the force of friction is 15N. With what speed will the sled emerge from the rough ice?

Answer 1

vf^2=Vi^2 + 2ad where a= force/mass, and it is negative here (it opposes motion). Solve for vf

Answer 2

a = F/m = 15N/45kg = 0.33 m/s^2 ,
vf^2 = vi^2 + 2ad
vf^2 = 3^2 + 2(0.33)(11)
vf^2 = 16.3
vf = 4m/s
is that correct?

Answer 3

a=F/m=15N/45kg=0.33 m/s^2
which in this case is Negative
vf^2=vi^2+2ad
vf^2=9+2(-.33)(11)
vf^2=9+(-7.26)
vf^2=1.74
vf=sqrt(1.74)
vf= 1.32 m/s