Abody of mass 25kg, moving at 3 ms per second on a rough horizontal floor brought to rest after sliding through a distance of 2.5m on the floor. calculate the coefficient of sliding friction [g = 10.0 ms per second per second]

You have to solve this by using the equations of motion:
u=3
v=0
s=2.5
a=?
v^2=u^2+2as
0=9+5s
Giving a=-1.8m/s^2

Then using the equation:
F=ma
F is the frictional force as there is no other force acting and its negative as its in the opposite direction to the direction of motion.

-F=25(-1.8)
F=45N

Then use the formula:
F=uR
Where u is the coefficient of friction, R is the normal force and F is the frictional force.

45=u(25g)
45=u(25*10)

Therefore, the coefficient of friction is 0.18

Hope that helps

Iwas going to ask that quesytion  sorry dude i dont  know : (    : /