A bag is gently dropped on a conveyor belt moving at a speed of $$2 \text{ m s}^{-1}$$. The coefficient of friction between the conveyor belt and bag is $$0.4$$. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion is: [Take $$g = 10 \text{ m s}^{-2}$$]

A bag is dropped on a conveyor belt moving at 2 m/s. We need to find the distance the bag travels on the belt during slipping.

The bag is dropped gently (initial velocity = 0) onto a belt moving at $$v = 2$$ m/s, and friction accelerates the bag until it reaches belt speed. The friction force on the bag is $$\mu m g$$, so the acceleration of the bag is $$a = \mu g = 0.4 \times 10 = 4 \text{ m/s}^2$$. The bag reaches the belt speed when $$v = a t$$, which gives $$2 = 4t \implies t = 0.5 \text{ s}$$.

During this time, the distance moved by the bag in the ground frame is $$s_{\text{bag}} = \frac{1}{2} a t^2 = \frac{1}{2} \times 4 \times 0.25 = 0.5 \text{ m}$$, and the belt moves $$s_{\text{belt}} = v \times t = 2 \times 0.5 = 1 \text{ m}$$. Therefore, the distance the bag travels relative to the belt while slipping is $$s_{\text{belt}} - s_{\text{bag}} = 1 - 0.5 = 0.5 \text{ m}$$.

The correct answer is Option B: $$0.5 \text{ m}$$.