[Physics] Newtonian mechanics, block on conveyor belt

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The coefficient of friction $\mu$ between the block and the conveyor belt is $0.3$ and the belt is moving with a speed of $6m/s$ when a block is dropped onto it from rest. Find the time when the relative motion between them stop.

My take:
When the block is dropped onto the belt friction would act on the block in the right hand side providing an acceleration of $\mu g$= $3m/s^2$ while an opposite friction would act on the conveyor belt giving it the same acceleration in the left side.

Writing the equation of motions for the block and belt;

FOR BLOCK :

$$v_{block}=at$$ where $ v_{block}$= final velocity; $a$=acceleration produced; $t$= time taken for that acceleration to be produced.$$=v_{block}=3t$$

For the BELT:

$$v_{belt}=6-(3)t$$

Now for the relative motion to stop between them

$$v_{belt}=v_{block}$$
$$6=6t$$
$$\Rightarrow t=1sec$$

But the answer according to the question is $t=2sec$

Can anybody please tell me where I did the question wrong?

Best Answer

Relative motion between block and belt will stop when speeds of block and belt become same. So, after finding the acceleration of the block due to friction, you should find the time taken for it to reach the speed of the belt. Note that, speed of the belt, is constant.

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