A bag is gently dropped on a conveyor belt moving at a speed of 2 m / s. 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 m / s⁻² ]

Solution

Physics sir · 4 steps

Friction is the only horizontal force, so the bag speeds up at μ g until it matches the belt.

1. Acceleration from friction. The bag starts at rest and the belt slides under it, so kinetic friction drags the bag forward. Its acceleration is a = μ g = 0.4 × 10 = 4 m s⁻².
2. When slipping stops. Slipping ends when the bag reaches the belt speed of 2 m s⁻¹. That takes t = 2/4 = 0.5 s.

   s = v²/2a = 2²/(2 × 4) = 0.5 m

3. Check from belt. In the belt's frame the bag starts at 2 m s⁻¹ backwards and slows down at 4 m s⁻². It slides 2²/(2 × 4) = 0.5 m on the belt, the same answer, because the belt moves 1 m while the bag moves 0.5 m.
4. So. The bag travels 0.5 m while it slips, which is option (b).

The idea to keep. For a body placed on a moving belt, friction gives acceleration μ g until the speeds match. The slipping distance is v²/(2μ g), both on the ground and relative to the belt.

Answer from the NTA final answer key. Solution by PrepWiser.