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 determine the force required to keep the conveyor belt moving at a constant speed and the power delivered by the motor.

### Step 1: Understanding the situation
The conveyor belt is moving at a constant speed of \( v = 2 \, \text{m/s} \), and sand is falling onto it at a rate of \( \frac{dm}{dt} = 5 \, \text{kg/s} \).

### Step 2: Calculate the momentum change
When the sand falls onto the conveyor belt, it has an initial velocity of \( 0 \, \text{m/s} \) (relative to the belt) and then moves with the belt at \( 2 \, \text{m/s} \). The change in momentum (\( \Delta p \)) for the sand falling onto the belt can be calculated as follows:

\[ \Delta p = m \cdot v \]

For \( 1 \, \text{s} \), the mass of sand falling is \( 5 \, \text{kg} \), and it gains a velocity of \( 2 \, \text{m/s} \):

\[ \Delta p = 5 \, \text{kg} \cdot 2 \, \text{m/s} = 10 \, \text{kg m/s} \]

### Step 3: Calculate the force required
The force (\( F \)) required to keep the conveyor belt moving can be calculated using the rate of change of momentum:

\[ F = \frac{d(\Delta p)}{dt} = \frac{10 \, \text{kg m/s}}{1 \, \text{s}} = 10 \, \text{N} \]

### Step 4: Calculate the power delivered by the motor
Power (\( P \)) can be calculated using the formula:

\[ P = F \cdot v \]

Substituting the values we have:

\[ P = 10 \, \text{N} \cdot 2 \, \text{m/s} = 20 \, \text{W} \]

### Final Results
- The force required to keep the belt moving is \( 10 \, \text{N} \).
- The power delivered by the motor is \( 20 \, \text{W} \).