Sand drops from a stationary hopper at the rate of `5kg//s` on to a conveyor belt moving with a constant speed of `2m//s`. What is the force required to keep the belt moving and what is the power delivered by the motor, moving the belt?

Explanation

To solve the problem, we need to find the force required to keep the conveyor belt moving and the power delivered by the motor. Let's break it down step by step.

### Step 1: Identify the given data
- Rate of sand dropping (dm/dt) = 5 kg/s
- Speed of the conveyor belt (v) = 2 m/s

### Step 2: Calculate the force required to keep the belt moving
According to Newton's second law, the force required to keep the conveyor belt moving can be calculated using the formula:

\[ F = \frac{d}{dt}(mv) \]

Where:
- \( m \) is the mass of sand falling per unit time (dm/dt)
- \( v \) is the velocity of the conveyor belt

Since the mass flow rate is constant, we can express the force as:

\[ F = v \cdot \frac{dm}{dt} \]

Substituting the values we have:

\[ F = 2 \, \text{m/s} \cdot 5 \, \text{kg/s} \]

Calculating this gives:

\[ F = 10 \, \text{N} \]

### Step 3: Calculate the power delivered by the motor
Power (P) can be calculated using the formula:

\[ P = F \cdot v \]

Where:
- \( F \) is the force calculated in the previous step
- \( v \) is the speed of the conveyor belt

Substituting the values:

\[ P = 10 \, \text{N} \cdot 2 \, \text{m/s} \]

Calculating this gives:

\[ P = 20 \, \text{W} \]

### Final Answers
- The force required to keep the belt moving is **10 Newtons**.
- The power delivered by the motor is **20 Watts**.

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