# If an object is moving at 100 m/s over a surface with a kinetic friction coefficient of u_k=5 /g, how much time will it take for the object to stop moving?

Jan 29, 2017

20 s

#### Explanation:

the initial kinetic energy will be dissipated by the friction of the object that is proportional to his weight, to the friction coefficient (fa) and to the length of the path (s):
$\frac{1}{2} m {v}^{2} = m g f a s$
but th e mass is the same and fa = 5/g so you can write
$\frac{1}{2} {v}^{2} = 5 s$
$s = {v}^{2} / 10$
$s = \frac{10000 {\left(\frac{m}{s}\right)}^{2}}{10 \left(\frac{m}{s} ^ 2\right)}$
s= 1000 m.
From the equations of accelerated motion
v° = at
s= -1/2 a t^2 + v°t
obtain "a" from the first equatinon and put it in the second
s= -1/2 (v°)/t t^2 +v°t = 1/2 v° t
 t = 2 s/( v° )= 2000m /(100 m/s) = 20 s.

This is only ipotetic, because the loss of energy in movement, especially at high velocity, depends by the force of resistance of air :
$F = f C x A {v}^{2}$
CX= aerodinamic coefficient
A = superficial area against motion direction
f= coefficient that takes account of temperature and air density...