Question

Sand from a stationary hopper falls onto a moving conveyor belt at a rate of 5.00 kg/s as shown in the figure. The conveyor belt is supported by frictionless rollers and moves at a constant speed of 0.750 m/s under the action of a constant horizontal external force $F_{ext}$ supplied by the motor that drives the belt.

• A. The force of friction exerted by the belt on the sand is 3.75 N.
• B. The external force $F_{ext}$ is 3.75 N.
• C. The work done by $F_{ext}$ in 1 sec is 2.81 J.
• D. The kinetic energy acquired by the falling sand each second due to the change in its horizontal motion is 1.41 J.

Answer 1

Answer: (A), (B), (C), (D)

A, B, C, D

Answer 2

The problem describes sand falling onto a moving conveyor belt. We need to analyze the forces, work done, and kinetic energy involved.

Given data:

• Rate of sand falling onto the belt, $\frac{dm}{dt} = 5.00 \text{ kg/s}$
• Constant speed of the conveyor belt, $v = 0.750 \text{ m/s}$

Let's analyze each option:

(A) The force of friction exerted by the belt on the sand is 3.75 N. When sand falls onto the belt, it initially has no horizontal velocity. To acquire the belt's speed, the belt must exert a forward frictional force on the sand. This force accelerates the incoming mass of sand. The force required to accelerate the sand to the belt's speed is given by the rate of change of momentum of the sand: $F_{friction\_on\_sand} = v \frac{dm}{dt}$ Substituting the given values: $F_{friction\_on\_sand} = (0.750 \text{ m/s}) \times (5.00 \text{ kg/s}) = 3.75 \text{ N}$ So, option (A) is correct.

(B) The external force $F_{ext}$ is 3.75 N. By Newton's third law, the sand exerts an equal and opposite frictional force on the belt, tending to slow it down. To maintain the constant speed of the belt, the external force $F_{ext}$ supplied by the motor must counteract this retarding frictional force. Therefore, $F_{ext} = F_{friction\_on\_belt} = F_{friction\_on\_sand} = 3.75 \text{ N}$. Alternatively, for a system where mass is continuously added and moves at a constant velocity, the external force required is $F_{ext} = v \frac{dm}{dt}$. $F_{ext} = (0.750 \text{ m/s}) \times (5.00 \text{ kg/s}) = 3.75 \text{ N}$ So, option (B) is correct.

(C) The work done by $F_{ext}$ in 1 sec is 2.81 J. Work done by a constant force is $W = F \times d$. In 1 second, the displacement of the belt (and thus the point of application of $F_{ext}$) is: $d = v \times t = (0.750 \text{ m/s}) \times (1 \text{ s}) = 0.750 \text{ m}$ The work done by $F_{ext}$ in 1 second is: $W_{ext} = F_{ext} \times d = (3.75 \text{ N}) \times (0.750 \text{ m}) = 2.8125 \text{ J}$ Rounding to two decimal places, $W_{ext} \approx 2.81 \text{ J}$. So, option (C) is correct.

(D) The kinetic energy acquired by the falling sand each second due to the change in its horizontal motion is 1.41 J. In 1 second, the mass of sand that falls onto the belt is: $m_{sand} = \frac{dm}{dt} \times 1 \text{ s} = (5.00 \text{ kg/s}) \times (1 \text{ s}) = 5.00 \text{ kg}$ This sand acquires a horizontal velocity equal to the belt's speed, $v = 0.750 \text{ m/s}$. The kinetic energy acquired by this sand is: $KE_{sand} = \frac{1}{2} m_{sand} v^2 = \frac{1}{2} \times (5.00 \text{ kg}) \times (0.750 \text{ m/s})^2$ $KE_{sand} = \frac{1}{2} \times 5.00 \times 0.5625 = 2.50 \times 0.5625 = 1.40625 \text{ J}$ Rounding to two decimal places, $KE_{sand} \approx 1.41 \text{ J}$. So, option (D) is correct.

All four options are correct.

Explanation of the solution:

1. Force on sand (A): The belt exerts a frictional force on the incoming sand to accelerate it to the belt's speed. This force is calculated as $F = v \frac{dm}{dt}$.
2. External force (B): To maintain constant belt speed, the external force must balance the backward frictional force exerted by the sand on the belt (which is equal in magnitude to the force calculated in A). Thus, $F_{ext} = v \frac{dm}{dt}$.
3. Work done by $F_{ext}$ (C): Work done is force times displacement. In 1 second, the belt moves a distance $d = v \times 1 \text{ s}$. The work done is $W_{ext} = F_{ext} \times d$.
4. Kinetic energy of sand (D): The mass of sand accumulated in 1 second is $\frac{dm}{dt} \times 1 \text{ s}$. This mass acquires kinetic energy $\frac{1}{2}mv^2$.