A box is sliding with a constant speed of 3.90 m/s in the +x-direction on a hori

Question

A box is sliding with a constant speed of 3.90 m/s in the +x-direction on a horizontal, frictionless surface. At x = 0 the box encounters a rough patch of the surface, and then the surface becomes even rougher. Between x = 0 and x = 2.00 m, the coefficient of kinetic friction between the box and the surface is 0.200; between x = 2.00 m and x = 4.00 m, it is 0.400. What is the x-coordinate of the point where the box comes to rest? Express your answer with the appropriate units. How much time does it take the box to come to rest after it first encounters the rough patch at x = 0? Express your answer with the appropriate units.

Explanation / Answer

we know that the force caused by the friccion is F=Nf where N is the normal force of the object and f is the coeficent of friction, so in this case the normal force is N=mg where m is the mass of the object and 9 is the gravity acceleration, g=9.81 m/s^2, so the force of firction is F=mgf and the acelerate that this object feels is a=gf in the -x direction, so we can know use cinematic equation to solve the problem:

first we got to know where the object is going to rest, so we use this:

vf^2=vi^2+2*x*a, where vf is the final speed, vi is the inicial speed (in this case 3.9 m/s) x is the distance and a is the desacelerating of the object.

first we evaluate the object traveling from 0 to 2 m: vf^2=3.9^2+2*2*(-9.81*0.2), and vf= 2.7133 m/2 this is the speed of the box when it pass the 2 m, if the speed was only negative, the box should stoped before the 2 m

next we tho the same, but in this case we want to know where te vf is 0 so we have this:

0=2.7133^2+2*x*(-9.81*0.4), and x results: 0.93807, so the object is going to rest before it pased the 4 m,

a) x=2+0.93807=2.93807 m from where it start

and for the b part, we calculate the time that it did everything, so form the first interval:

d=v*t+(1/2)a*t^2, that is 2=(3.9*t)-(1/2)(9.81*0.2*t^2), t=3.37 seg

and for the second, we use the same equation:

0.938073=2.7133*t-(1/2)*(9.81*0.4*t^2), t=0.755 seg

b) so the total time is t=0.755+3.37= 4.125 seg

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