Sand is being dropped on a conveyor belt at the rate of `Mkg//s` . The force necessary to kept the belt moving with a constant with a constant velocity of `vm//s` will be.

Options

A. Mv newton
B. 2 Mv newton
C. `(Mv)/(2)` newton
D. zero

Explanation

To solve the problem, we need to determine the force necessary to keep a conveyor belt moving at a constant velocity when sand is being dropped on it at a rate of \( M \) kg/s.

### Step-by-Step Solution:

1. **Understand the Problem**:
- We have a conveyor belt moving with a constant velocity \( v \) m/s.
- Sand is being dropped onto the belt at a rate of \( M \) kg/s.

2. **Apply Newton's Second Law**:
- According to Newton's second law of motion, the force \( F \) is equal to the rate of change of momentum \( \frac{dp}{dt} \).
- The momentum \( p \) of an object is given by \( p = mv \), where \( m \) is the mass and \( v \) is the velocity.

3. **Differentiate Momentum**:
- The change in momentum with respect to time can be expressed as:
\[ F = \frac{dp}{dt} = \frac{d(mv)}{dt} \]
- Using the product rule of differentiation, we have:
\[ F = v \frac{dm}{dt} + m \frac{dv}{dt} \]

4. **Identify Constants**:
- Since the conveyor belt moves with a constant velocity, \( \frac{dv}{dt} = 0 \).
- Therefore, the second term \( m \frac{dv}{dt} \) becomes zero.

5. **Substituting Values**:
- We know that \( \frac{dm}{dt} = M \) (the rate at which sand is dropped).
- Substituting these values into the equation gives:
\[ F = v \cdot M + 0 \]
- Thus, the force required to keep the conveyor belt moving at a constant velocity is:
\[ F = Mv \]

### Final Answer:
The force necessary to keep the belt moving with a constant velocity \( v \) m/s is \( F = Mv \).

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