Question 1182828
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At the instant a truck is travelling along a level pavement at 20 m/s, a wooden crate falls from the truck. 
If the crate travels 40 meters before coming to rest, what is the coefficient of kinetic friction between 
the crate and the road?
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<pre>
In this Physics problem, we are given that a wooden crate slides along the level pavement,
having the initial speed of 20 m/s, and stops after traveling 40 meters.


The simplest way to solve the problem is to apply the Law of Conservation mechanical energy.


From one side, we have the kinetic energy  {{{(mv^2)/2}}}  of the sliding crate,

where  "m"  is its mass and "v"  is its initial speed.



From the other side, we have the work of the friction force  F*d,  where F is the friction force  k*(m*g),

k is the kinematic friction coefficient,  g is the gravity acceleration and  "d"  is the the breaking distance (given).


The Conservation Law of mechanical energy allows us to write this equation


    {{{(m*v^2)/2}}} = k*m*g*d.


First, we cancel the mass in both sides and then we get the formula for the friction coefficient


    k = {{{(v^2)/(2*g*d)}}} = {{{20^2/(2*10*40)}}} = {{{1/2}}} = 0.5.


<U>ANSWER</U>.  Under given conditions, the kinematic friction coefficient is  0.5.
</pre>

Solved.