1. Initial situation

A bag is dropped gently on a conveyor belt moving with speed $v = 2\,\text{m/s}.$

Since the bag is dropped gently, its initial horizontal speed is $u = 0.$

So relative to the ground, the bag starts from rest and the belt moves beneath it at $2\,\text{m/s}$.

2. Force acting during slipping

While the bag is slipping, friction acts on the bag in the direction of motion of the belt.

Magnitude of friction: $f = \mu mg$ where $\mu = 0.4, \quad g = 10\,\text{m/s}^2.$

Thus the acceleration of the bag is $a = \frac{f}{m} = \mu g = 0.4 \times 10 = 4\,\text{m/s}^2.$

3. Condition for slipping to stop

Slipping stops when the bag attains the same speed as the belt, i.e. $v_f = 2\,\text{m/s}.$

Using $v_f^2 = u^2 + 2as,$ with $u=0, \quad v_f=2, \quad a=4,$ we get $2^2 = 0 + 2(4)s$ $4 = 8s$ $s = 0.5\,\text{m}.$

4. Answer

Therefore, the distance travelled by the bag during slipping is $\boxed{0.5\,\text{m}}.$

So the correct option is B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They match.