a body of mass 25kg moving at 3m/s on a rough horizontal floor is brought to rest after sliding through a distance of 2.5m on the floor. calculate the coefficient of sliding friction.

t is time to stop

a = 3/t

d = (1/2)a t^2
2.5 = .5 (3/t) t^2
2.5 = 1.5 t
t = 1.67 seconds to stop
then
a = 3/t = 1.8 m/s^2
F = ma
F = 25 * 1.8 = 45 Newtons
weight = 25g = 245 Newtons

coef of friction = F/weight = 45/245
= .184

Thank you so much for helping me solve this problem

f/mg =45\25*10=45/250=0.18

bruh,its practically the same damn thing. .184 is 0.18, you just approximated.

a = v2−u22a=02−322×2.5 = 1.8m/s2

F = ma = 25 x 1.8 = 45N

but μ=FR=4525×10=0.18

not correct

not f\weight

Partially wrong

Calculate the coefficient sliding giving that g = 10m/s²s².after the answer DAT what they said we should find

A body of mass,25kg at 3m/s on a rough horizontal floor is bought to fest after sliding though a distance of 2.50m on the floor calculate the co-efficient of the sliding rotation