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A horizontal conveyor belt is moving at ...

A horizontal conveyor belt is moving at a speed of 5m/s. A box of mass 20kg is gently placed on this belt. Box first slips on the belt and finally comes to rest with respect to belt. If box takes time 0.1 s to stop slipping on the belt, then what will be the distance travelled by the box during this interval?

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To solve the problem step-by-step, we can follow these steps: ### Step 1: Understand the Initial Conditions - The conveyor belt is moving at a speed of \( v_b = 5 \, \text{m/s} \). - The box has a mass of \( m = 20 \, \text{kg} \). - The box is placed on the belt and initially has a velocity of \( v_{box, initial} = 0 \, \text{m/s} \) relative to the ground. - The box takes \( t = 0.1 \, \text{s} \) to come to rest with respect to the belt. ...
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Knowledge Check

  • A conveyor belt is moving at a constant speed of 2m//s . A box is grenty dropped on it. Th ecoefficient of friction between them is mu=0.5 . The distance that the box will move relative to belt before coming to rest on it taking g=10ms^(-2) is:

    A
    `1.2m`
    B
    `0.6m`
    C
    zero
    D
    `0.4m`

  • A conveyer belt is moving at a constant speed of 2 ms^(-1) A box is gently dropped on it. The coefficient of friction between them is mu = 0.5. The distance that the box will move relative to belt before coming to rest on it, taking g= 10 ms^(-2) , is

    A
    `0.4 m`
    B
    1.2 m
    C
    0.6m
    D
    zero

  • A long horizontal belt is moving from left to right with a uniform speed 2 m/s. There are two ink marks A and B on the belt 60 m apart. An insect runs on the belt to the fro between A and B such that its speed releatice to belt is constant and equals 4 m/s. When the insect is moving on the belt in the directoin of motion of the belt, its speed as observed by a person standing on ground will be:

    A
    6 m/s
    B
    2 m/s
    C
    1.5 m/s
    D
    4 m/s

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