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3 years ago

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A box is dropped onto a conveyor belt moving at 3.2 m/s. If the coefficient of friction between the box and the belt is 0.28, ho

w long will it take (in s) before the box moves without slipping

1 answer:

Lemur [1.5K]3 years ago

6 0

Answer:

t = 1.16 s.

Explanation:

Given,

speed of conveyor belt, v = 3.2 m/s

coefficient of friction,f = 0.28

Using newton second law

f = ma

and we also know that frictional force

f = μ N

f = μ m g

equating both the force equation

a = μ g

a = 0.28 x 9.81

a = 2.75 m/s²

Using Kinematic equation

v = u + at

3.2 = 0 + 2.75 x t

t = 1.16 s.

Time taken by the box to move without slipping is 1.16 s.

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