Sand is droped vetically downward at the constant rate `mu kg//s` on a conveyor belt which is moving horizontally with velocity `'v'`. Assume that sant particles comes to rest with respect to belt immediately after landing.
The rate which work is done by friction force on sand is

Options

A. `(muv^(2))/(2)`
B. `-muv^(2)`
C. `-(muv^(2))/(2)`
D. `-(3)/(2)muv^(2)`

Explanation

To solve the problem, we need to determine the rate at which work is done by the friction force on the sand as it lands on the conveyor belt. Here’s the step-by-step solution:

### Step 1: Understand the system
- Sand is dropped vertically at a constant rate of `μ kg/s` onto a conveyor belt moving horizontally with a velocity `v`.
- When the sand particles land on the belt, they come to rest relative to the belt immediately.

### Step 2: Determine the mass on the conveyor belt
- At any time `t`, the mass of sand that has landed on the conveyor belt is given by:
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