Question

A stone is pushed over the floor with a velocity of 2m/s and stops after sliding a distance of 5m finds its coefficient of friction

Answer 1

u = 2m/s

v= 0

s = 5m

k(coeff. of friction) = ?

Apply
$v {}^{2} = \: u {}^{2} - 2as$

0 = 4 - 10a

a = 0.4m/s^2

Now , Friction force = ma

kmg = ma

k = a/g

k = 0.4 / 0 = 0.04

k = 0.04



Thanks