Packages having a mass of 6 kgkg slide down a smooth chute and land horizontally with a speed of 3 m/sm/s on the surface of a conveyor belt. If the coefficient of kinetic friction between the belt and a package is

t = 1.02 s

Explanation:

The computation of the time required is shown below:

The package speed for belt is

= 3 -  1

= 2 m/s

Moreover, the decelerative force would be acted on the block i.e u.m.g

So, the decelerative produced

= 0.2 × 9.81

= 1.962 m/s^2

And, final velocity = 0

v = u - at

here

V = 0 = final velocity

 u = 2 m/s

so,

0 = 2 - 1.962 × t

t = 1.02 s