Question

A conveyor belt is moving at a constant speed of $2\, ms^{-1}.$ A box is gently dropped on it. The coefficient of friction between them is $\mu = 0.5$. The distance that the box will move relative to belt before coming to rest on it, taking $g = 10\, ms^{-2}$, is

• A. 0.4 m
• B. 1.2 m
• C. 0.6 m
• D. Zero

Answer 1

The Correct Option is A

To solve this problem, we need to find out how far the box travels on the conveyor belt before it stops moving relative to the belt. The box is initially dropped onto a moving conveyor belt with a speed of 2\, \text{ms}^{-1}.

Since the box is initially at rest with respect to an observer on the ground, and the belt is moving, there will be a kinetic friction force acting between the box and the belt that will accelerate the box until it moves at the same speed as the belt.

The kinetic friction force f_k is given by:

f_k = \mu \cdot m \cdot g

Where:

• \mu = 0.5 (coefficient of friction)
• m is the mass of the box (which will cancel out later)
• g = 10\, \text{ms}^{-2} (acceleration due to gravity)

The box will experience an acceleration a due to this frictional force, which can be given by Newton's second law:

f_k = m \cdot a \Rightarrow a = \frac{f_k}{m} = \mu \cdot g = 0.5 \cdot 10 = 5\, \text{ms}^{-2}

We use the formula for the distance covered under constant acceleration, starting from rest, to find the distance d:

v^2 = u^2 + 2ad

Where:

• u = 0 \, \text{ms}^{-1} (initial relative velocity between the box and the belt)
• v = 2 \, \text{ms}^{-1} (final relative velocity between the box and the ground is zero when it's moving with the belt)
• a = 5\, \text{ms}^{-2} (acceleration due to friction)

Substitute into the equation:

0^2 = (2\, \text{ms}^{-1})^2 + 2 \cdot (-5 \, \text{ms}^{-2}) \cdot d

0 = 4 - 10d

Solving for d:

10d = 4 \Rightarrow d = \frac{4}{10} = 0.4\, \text{m}

Thus, the distance the box will move relative to the belt before coming to rest on it is 0.4 meters.