Question:
Sand falls on a conveyor belt at the rate of 1.5 kg/s. If the belt is moving with a constant speed of 7 m/s, the power needed to keep the conveyor belt running is:
Sand falls on a conveyor belt at the rate of 1.5 kg/s. If the belt is moving with a constant speed of 7 m/s, the power needed to keep the conveyor belt running is:
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The power required to move an object on a conveyor belt is equal to the rate at which kinetic energy is being added to the object.
Updated On: Nov 18, 2025
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Solution and Explanation
Step 1: Understand the problem.
The power required to keep the conveyor belt moving is related to the rate of change of kinetic energy of the sand. The kinetic energy per unit time (power) is given by the formula: \[ P = \frac{d}{dt} \left( \frac{1}{2} m v^2 \right) \] where \( m = 1.5 \, \text{kg/s} \) is the mass flow rate of sand, and \( v = 7 \, \text{m/s} \) is the velocity of the conveyor belt.
Step 2: Apply the formula.
The power required is: \[ P = \frac{1}{2} \times 1.5 \times 7^2 = \frac{1}{2} \times 1.5 \times 49 = 36.75 \, \text{Watts} \] Step 3: Conclusion.
Thus, the power required to keep the conveyor belt running is 36.75 Watts.
The power required to keep the conveyor belt moving is related to the rate of change of kinetic energy of the sand. The kinetic energy per unit time (power) is given by the formula: \[ P = \frac{d}{dt} \left( \frac{1}{2} m v^2 \right) \] where \( m = 1.5 \, \text{kg/s} \) is the mass flow rate of sand, and \( v = 7 \, \text{m/s} \) is the velocity of the conveyor belt.
Step 2: Apply the formula.
The power required is: \[ P = \frac{1}{2} \times 1.5 \times 7^2 = \frac{1}{2} \times 1.5 \times 49 = 36.75 \, \text{Watts} \] Step 3: Conclusion.
Thus, the power required to keep the conveyor belt running is 36.75 Watts.
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