Displacement of block upto that instant. (g= $10 m / s^{2}$ ).

6 m

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2 m

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4 m

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5 m

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Solution

The correct option is C 6 m
We have
$F = μ m g$
Retardation on belt is given by
$a = \frac{F}{m} = \frac{μ m g}{m} = μ g$
$a = 0.3 \times 10 = 3$
Now,
$v^{2} = u^{2} + 2 a S 0 = 6^{2} - 2 \times 3 \times S S = 6 m$

Option A is correct.

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