# [Physics] Newtonian mechanics, block on conveyor belt

free-body-diagramhomework-and-exercisesnewtonian-mechanics

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?