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A factory robot drops a 10 kg computer onto a conveyor belt running at 3.1 m/s. The materials are such that
?s = 0.50 and ?k = 0.30 between the belt and the computer. How far is the computer dragged before it is riding smoothly on the belt?
A) 0.98 m
B) 3.0 m
C) 1.6 m
D) 2.3 m

Respuesta :

Answer:

Option (c) will be correct answer that is it will go 1.6 m

Explanation:

We have given that conveyor has the velocity u = 3.1 m/sec

Mass of the robot = 10 kg

static friction coefficient  = 0.5 and kinetic friction coefficient = 0.3

Acceleration due to gravity g = 9.8 [tex]m/sec^2[/tex]

Acceleration a = kinetic friction coefficient ×g = 0.3×9.8 = 2.94[tex]m/sec^2[/tex]

Now according to third equation of motion

[tex]v^2=u^2+2as[/tex]

Finally velocity of the conveyor will be zero

So [tex]0^2=3.1^2-2\times 2.94\times s[/tex]

s = 1.6 m

So option (c) is correct option

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