Question

A conveyor belt is moving horizontally with a velocity of \( 2 \, \text{m s}^{-1} \). If a body of mass 10 kg is kept on it, then the distance travelled by the body before coming to rest is (The coefficient of kinetic friction between the belt and the body is 0.2 and acceleration due to gravity is \( 10 \, \text{m s}^{-2} \))

• A. 4 m
• B. 0 m
• C. 1 m
• D. 2 m

Hint:

When a body is placed on a moving conveyor belt, kinetic friction acts to bring the body to the same velocity as the belt (i.e., to rest relative to the belt). Friction force \(f_k = \mu_k N = \mu_k mg\). Acceleration of the body \(a = f_k/m = \mu_k g\). Use kinematic equations (\(v=u+at\), \(s=ut+\frac{1}{2}at^2\), \(v^2=u^2+2as\)) to find the distance travelled until the body's velocity matches the belt's velocity. "Coming to rest" here likely means relative to the belt.

Answer 1

The Correct Option is C

The question asks for the distance travelled by the body before coming to rest.
This implies the body comes to rest relative to the ground, if initially it was not moving with the belt.
However, if the body is "kept on it", it implies its initial velocity relative to the belt is zero, or its initial velocity relative to ground is zero and the belt then imparts motion.
The phrasing "before coming to rest" usually means relative to the reference frame in which its initial velocity was non-zero and final velocity is zero.
Let's assume the body is placed on the moving belt.
Initially, the body is at rest relative to the ground (or has some other velocity).
The belt is moving at \( v_b = 2 \, \text{m s}^{-1} \).
If the body is placed on the belt, it will experience a kinetic friction force that tries to make it move with the belt.
This friction will accelerate the body.
The body will come to rest relative to the belt when its velocity reaches \( 2 \, \text{m s}^{-1} \).
The question phrasing "coming to rest" seems to imply relative to the ground, meaning its velocity becomes 0.
This can only happen if the body was initially thrown onto the belt with a velocity different from the belt's velocity, and friction opposes its relative motion.
Interpretation: The body is placed gently on the belt.
Its initial velocity \(u_{body,ground}=0\).
The belt moves at \(v_{belt}=2 \text{ m/s}\).
Friction force \( f_k = \mu_k N \).
Here \( N = mg \).
\( f_k = \mu_k mg = 0.
2 \times 10 \text{ kg} \times 10 \text{ m/s}^2 = 20 \text{ N} \).
This force accelerates the body: \( a_{body} = f_k / m_{body} = \mu_k g = 0.
2 \times 10 = 2 \text{ m/s}^2 \).
The body accelerates until its velocity equals the belt's velocity.
Let \(v = u + at \implies 2 = 0 + 2t \implies t=1 \text{ s}\).
Distance travelled by the body relative to the ground during this time: \( s = ut + \frac{1}{2}at^2 = 0 \cdot 1 + \frac{1}{2}(2)(1)^2 = 1 \text{ m} \).
At this point, the body is moving with the belt at \(2 \text{ m/s}\) and there is no more relative motion, so kinetic friction ceases (or static friction might act if there are other forces).
The question wording "before coming to rest" must mean "before coming to rest relative to the belt".
Alternative Interpretation: The body is moving on the belt and the belt is what causes it to stop.
This implies the body has an initial velocity relative to the ground, and the belt opposes this.
This is less likely for "kept on it".
If the question meant: "A body is on a conveyor belt that is initially moving at \(2 \text{ m/s}\).
The belt then stops.
What distance does the body travel on the belt before coming to rest?" Then initial velocity of body \(u=2 \text{ m/s}\).
Friction causes deceleration \(a = -\mu_k g = -2 \text{ m/s}^2\).
Using \( v^2 = u^2 + 2as \): \( 0^2 = 2^2 + 2(-2)s \implies 0 = 4 - 4s \implies 4s=4 \implies s=1 \text{ m} \).
This interpretation also gives 1 m.
Let's consider the frame of reference of the belt.
Initial velocity of body relative to belt \( u_{rel} = u_{body,ground} - v_{belt} \).
If body is gently placed, \( u_{body,ground} = 0 \), so \( u_{rel} = -2 \text{ m/s} \).
The friction acts to reduce this relative velocity to 0.
Acceleration of body due to friction is \( a_f = \mu_k g = 2 \text{ m/s}^2 \) (in the direction of belt's motion).
In the belt's frame, the body experiences an acceleration of \( a_f \) relative to the ground, so its acceleration relative to the belt is \( a_{rel} = a_{body,ground} - a_{belt,ground} \).
If the belt moves at constant velocity, \( a_{belt,ground}=0 \).
So \( a_{rel} = a_{body,ground} = 2 \text{ m/s}^2 \) (in the direction of the belt's movement).
The body is initially "slipping backward" at \( -2 \text{ m/s} \) relative to the belt.
Friction accelerates it forward.
Distance travelled relative to the belt: \( v_{rel}^2 = u_{rel}^2 + 2 a_{f, in belt frame} s_{rel} \).
Final relative velocity is 0.
Initial relative velocity is -2 m/s.
The friction force acts on the body in the direction of the belt's motion.
So, it accelerates the body from 0 to 2 m/s relative to the ground.
Distance travelled by the body (relative to ground) = \(s\).
\(v^2 = u^2 + 2as \implies 2^2 = 0^2 + 2(2)s \implies 4 = 4s \implies s = 1 \text{ m}\).
This is the distance travelled by the body (on the ground) until it moves with the belt.
At this point, it is "at rest" relative to the belt.
The phrasing is a bit ambiguous, but 1m seems to be the consistent answer.
This matches option (3).