Question

A very long horizontal conveyer belt is always moving with constant speed of 10 m/s with the help of electric motor. A block of mass 2 kg is placed on the belt after giving it a horizontal velocity of 20 m/s in direction opposite to the velocity of the belt. Assume that after some time, rubbing between the belt and block stops due to friction between them. Let H₁ be the heat produced during the motion of block in direction opposite to the motion of the belt. Let H₂ be the heat produced during the motion of block in direction of motion of the belt. Choose the correct statement(s)

• A. H₁ = 400 J
• B. H₁ = 800 J
• C. H₂ = 100 J
• D. H₂ = 200 J

Answer 1

Answer: (B), (C)

H₁ = 800 J, H₂ = 100 J

Answer 2

Let the velocity of the conveyor belt be $v_b = +10$ m/s. The initial velocity of the block is $v_i = -20$ m/s (opposite to the belt's velocity). The mass of the block is $m = 2$ kg.

The friction force acting on the block is kinetic friction as long as there is relative motion. The magnitude of the kinetic friction force is $f_k = \mu_k N = \mu_k mg$, where $\mu_k$ is the coefficient of kinetic friction and $N=mg$ is the normal force.

The relative velocity of the block with respect to the belt is $v_{rel} = v_{block} - v_{belt}$.

Initially, $v_{rel, i} = -20 - 10 = -30$ m/s. The block is moving to the left relative to the belt.

The friction force on the block is in the direction opposite to the relative velocity of the block with respect to the belt, i.e., in the direction of the belt's velocity (positive direction).

The acceleration of the block is $a = \frac{f_k}{m} = \frac{\mu_k mg}{m} = \mu_k g$. This acceleration is constant and in the positive direction.

The velocity of the block at time $t$ is $v(t) = v_i + at = -20 + \mu_k g t$.

The block starts with $v = -20$ m/s and accelerates towards the positive direction. Its velocity increases from -20 m/s, passes through 0 m/s, and eventually reaches +10 m/s, at which point the relative velocity is zero ($10 - 10 = 0$), and the block moves with the belt.

Let's define the phases based on the direction of the block's motion relative to the ground.

Phase 1: The block moves in the direction opposite to the belt's motion. This happens when the block's velocity $v$ is negative. The velocity changes from -20 m/s to 0 m/s.

Let $t_1$ be the time when the block's velocity becomes 0 m/s.

$0 = -20 + \mu_k g t_1 \implies t_1 = \frac{20}{\mu_k g}$.

During this phase ($0 \le t \le t_1$), the block moves in the negative direction.

Let's calculate the heat produced during this phase, $H_1$. Heat produced by friction is $H = f_k \times |\Delta x_{rel}|$, where $|\Delta x_{rel}|$ is the magnitude of the relative displacement between the surfaces during the period when kinetic friction acts.

The relative velocity is $v_{rel}(t) = v(t) - v_b = (-20 + \mu_k g t) - 10 = -30 + \mu_k g t$.

The relative displacement during Phase 1 ($0 \le t \le t_1$) is $\Delta x_{rel, 1} = \int_0^{t_1} v_{rel}(t) dt = \int_0^{t_1} (-30 + \mu_k g t) dt = [-30t + \frac{1}{2} \mu_k g t^2]_0^{t_1}$.

Substitute $t_1 = \frac{20}{\mu_k g}$:

$\Delta x_{rel, 1} = -30(\frac{20}{\mu_k g}) + \frac{1}{2} \mu_k g (\frac{20}{\mu_k g})^2 = -\frac{600}{\mu_k g} + \frac{1}{2} \mu_k g \frac{400}{(\mu_k g)^2} = -\frac{600}{\mu_k g} + \frac{200}{\mu_k g} = -\frac{400}{\mu_k g}$.

The magnitude of the relative displacement is $|\Delta x_{rel, 1}| = \frac{400}{\mu_k g}$.

The heat produced $H_1 = f_k \times |\Delta x_{rel, 1}| = (\mu_k mg) \times \frac{400}{\mu_k g} = 400m$.

Given $m=2$ kg, $H_1 = 400 \times 2 = 800$ J.

Phase 2: The block moves in the direction of the belt's motion. This happens when the block's velocity $v$ is positive. The velocity changes from 0 m/s to +10 m/s.

Let $t_2$ be the time when the block's velocity becomes +10 m/s.

$10 = -20 + \mu_k g t_2 \implies t_2 = \frac{30}{\mu_k g}$.

This phase occurs during the time interval $[t_1, t_2]$.

During this phase ($t_1 \le t \le t_2$), the block moves in the positive direction.

The relative velocity is $v_{rel}(t) = -30 + \mu_k g t$.

At $t=t_1 = \frac{20}{\mu_k g}$, $v_{rel}(t_1) = -30 + \mu_k g (\frac{20}{\mu_k g}) = -30 + 20 = -10$ m/s.

At $t=t_2 = \frac{30}{\mu_k g}$, $v_{rel}(t_2) = -30 + \mu_k g (\frac{30}{\mu_k g}) = -30 + 30 = 0$ m/s.

The relative velocity is negative during this phase, meaning the block is still moving to the left relative to the belt.

The relative displacement during Phase 2 ($t_1 \le t \le t_2$) is $\Delta x_{rel, 2} = \int_{t_1}^{t_2} v_{rel}(t) dt = \int_{t_1}^{t_2} (-30 + \mu_k g t) dt = [-30t + \frac{1}{2} \mu_k g t^2]_{t_1}^{t_2}$.

Substitute $t_1 = \frac{20}{\mu_k g}$ and $t_2 = \frac{30}{\mu_k g}$:

$\Delta x_{rel, 2} = (-30(\frac{30}{\mu_k g}) + \frac{1}{2} \mu_k g (\frac{30}{\mu_k g})^2) - (-30(\frac{20}{\mu_k g}) + \frac{1}{2} \mu_k g (\frac{20}{\mu_k g})^2)$.

$\Delta x_{rel, 2} = (-\frac{900}{\mu_k g} + \frac{450}{\mu_k g}) - (-\frac{600}{\mu_k g} + \frac{200}{\mu_k g})$.

$\Delta x_{rel, 2} = (-\frac{450}{\mu_k g}) - (-\frac{400}{\mu_k g}) = -\frac{50}{\mu_k g}$.

The magnitude of the relative displacement is $|\Delta x_{rel, 2}| = \frac{50}{\mu_k g}$.

The heat produced $H_2 = f_k \times |\Delta x_{rel, 2}| = (\mu_k mg) \times \frac{50}{\mu_k g} = 50m$.

Given $m=2$ kg, $H_2 = 50 \times 2 = 100$ J.

So, $H_1 = 800$ J and $H_2 = 100$ J.