An elevator that can carry a maximum load of 1500 kg (elvator+ passengers) is moving up with a constant speed of 2 `ms^(-1).` The frictional force opposing the motion is 3000 N. Find the minmum power delvered by the motor to the elevator in watts as well as in horse power `(g=10ms^(-2))`

Explanation

To solve the problem, we need to calculate the minimum power delivered by the motor to the elevator. Here are the steps to find the solution:

### Step 1: Calculate the weight of the elevator and passengers
The total mass of the elevator and passengers is given as 1500 kg. The weight (W) can be calculated using the formula:
\[ W = m \cdot g \]
Where:
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