Posted by Lala on .
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?

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

Physics 
Lala,
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? 
Physics 
Kyle,
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