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