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A conveyor belt is moving at a constant speed of 2m/s. 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=10ms-2, is 

  • Option 1)

    1.2m

  • Option 2)

    0.6m

  • Option 3)

    Zero

  • Option 4)

    0.4m

 

Answers (1)

best_answer

As we discussed in concept

Friction -

Opposing Force which is parallel to the surface and opposite to direction of Relative Motion.

- wherein

Direction of friction is always opposite the direction of Relative Motion.

 

 Force of friction f= \mu mg

\therefore \therefore\:a=\frac{f}{m}\:=\:\frac{\mu mg}{m}\:=\: \mu g\:=\: 0.5\times 10\:=\:5\:ms^{-2}

Now v^{2}=u^{2}v^{2}=\mu ^{2}+2as\:=>\:v^{2}- \mu^{2}\:=\:2as

o^{2}-2^{2}=2(-5)\times s\:\:\:=>s=0.4\:mt

 


Option 1)

1.2m

This option is incorrect.

Option 2)

0.6m

This option is incorrect.

Option 3)

Zero

This option is incorrect.

Option 4)

0.4m

This option is correct.

Posted by

divya.saini

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