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]

Given:

Initial speed of the bag relative to the conveyor belt: v₀ = 2 m/s

Coefficient of friction between the conveyor belt and the bag: μ = 0.4

Acceleration due to gravity: g = 10 m/s²

Solution:

The frictional force acting on the bag will decelerate it until it matches the speed of the conveyor belt. The frictional force can be given by:

f_friction = μ m g

Where m is the mass of the bag (which will cancel out during the calculations).

The work done by the frictional force to stop the slipping is equal to the change in kinetic energy of the bag. Initially, the bag has a kinetic energy of:

KE = ½ m v₀²

The work done by friction is:

W = f_friction × d

Where d is the distance traveled during slipping. The work done by friction will equal the change in kinetic energy, so:

μ m g × d = ½ m v₀²

The mass m cancels out from both sides:

μ g × d = ½ v₀²

Now, solving for d:

d = (v₀²) / (2 μ g)

Substitute the known values:

d = (2²) / (2 × 0.4 × 10)

Simplifying:

d = 4 / 8 d = 0.5 m

Final Answer:

The distance traveled by the bag on the belt during the slipping motion is: 0.5 m