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 $F_{e x t}$ supplied by the motor that drives the belt.

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The force of friction exerted by the belt on the sand is 3.75 N.

The kinetic energy acquired by the falling sand each second due to the change in its horizontal motion is 1.41 J.

The external force $F_{e x t}$ is 3.75 N.

The work done by $F_{e x t}$ in 1 sec is 2.81 J.

answer is A, B, C, D.

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Detailed Solution

(ABCD)
$\frac{d m}{d t} = 5 k g / s v_{C} = 0.75 m / s F_{e x t} - f = 0 F_{e x t} = f \vec{f} = \frac{m d \vec{v}}{d t} - {\vec{v}}_{r e l} \frac{d m}{d t} f \hat{i} = - (0 - v_{C} \hat{i}) \frac{d m}{d t} f = 0.75 \times 5 f = 3.75 N W = \int_{i}^{f} F_{e x t} d S = 3.75 (S) = 3.75 (v_{C} t) W = 3.75 (0.75 \times 1) = 2.81 J$
$\frac{d k}{d t} = \frac{1}{2} (\frac{d m}{d t}) v^{2} = \frac{1}{2} (5) \times {(0.75)}^{2} = 1.41 J$ .