Question

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/s²]

• A. 2 m
• B. 0.5 m
• C. 3.2 m
• D. 0.8 ms

Answer 1

The Correct Option is B

 To solve the problem of finding the distance traveled by the bag during its slipping motion on the conveyor belt, we need to understand the mechanics involved and apply the relevant physics concepts step by step.

When the bag is dropped on the conveyor belt, it initially has zero velocity relative to the ground but is in contact with a belt moving at 2 m/s. The friction between the belt and bag will cause the bag to eventually match the belt's speed, but initially, it will slip.

The force of friction, which is the decelerating force here, is what acts to accelerate the bag to the conveyor's speed. This frictional force can be calculated using the formula:

\(F_{\text{friction}} = \mu \cdot m \cdot g\)

where:

• \(\mu = 0.4\) is the coefficient of friction,
• \(m\) is the mass of the bag (which ultimately cancels out), and
• \(g = 10 \, \text{m/s}^2\) is the acceleration due to gravity.

 

The acceleration \(a\) experienced by the bag due to friction is given by:

\(a = \mu \cdot g = 0.4 \times 10 \, \text{m/s}^2 = 4 \, \text{m/s}^2\)

Considering the case where the bag moves from a state of rest on the belt to reaching the belt's speed (slipping stops), we use the kinematic equation to find the distance \(s\) traveled during slipping:

\(v^2 = u^2 + 2as\)

where:

• \(v = 2 \, \text{m/s}\) is the final velocity (same as the belt),
• \(u = 0\) is the initial velocity of the bag relative to the belt.

 

Substituting the known values:

\((2)^2 = 0 + 2 \times 4 \times s\)

\(4 = 8s\)

Solving for \(s\), we get:

\(s = \frac{4}{8} = 0.5 \, \text{m}\)

Thus, the distance traveled by the bag on the belt during the slipping motion is 0.5 meters.