An elevator has full load capacity 800 kg. Frictional force on the elevator is 4000 N. If elevator is going down with constant speed 2.0 m/s, then the power applied by electric motor at full load is lose to : (1 HP = 746 W, g = 10 `m//s^2`)

Options

A. -10.72 HP
B. -32.17 HP
C. +10.72 HP
D. +32.17 HP

Explanation

To solve the problem step by step, we will follow the principles of physics related to forces and power.

### Step 1: Identify the forces acting on the elevator.
- The weight of the elevator (W) is given by the formula:
\[ W = m \cdot g \]
where \( m = 800 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \).

### Step 2: Calculate the weight of the elevator.
\[ W = 800 \, \text{kg} \times 10 \, \text{m/s}^2 = 8000 \, \text{N} \]

### Step 3: Write the equation of motion for the elevator.
Since the elevator is moving down with a constant speed, the net force acting on it is zero. The forces acting on the elevator are:
- Tension (T) acting upwards
- Frictional force (F) acting upwards
- Weight (W) acting downwards

The equation can be written as:
\[ T + F = W \]

### Step 4: Solve for the tension (T).
Rearranging the equation gives:
\[ T = W - F \]
Substituting the values:
\[ T = 8000 \, \text{N} - 4000 \, \text{N} = 4000 \, \text{N} \]

### Step 5: Calculate the power applied by the electric motor.
Power (P) is given by the formula:
\[ P = F \cdot v \]
In this case, the tension (T) is the force applied by the motor, and the velocity (v) is given as \( 2.0 \, \text{m/s} \). However, since the elevator is moving down and the tension is acting upwards, we take the angle between the force and velocity into account.

The angle between the tension and the velocity is \( 180^\circ \), so:
\[ P = T \cdot v \cdot \cos(180^\circ) = T \cdot v \cdot (-1) \]
Substituting the values:
\[ P = 4000 \, \text{N} \cdot 2.0 \, \text{m/s} \cdot (-1) = -8000 \, \text{W} \]

### Step 6: Convert the power from watts to horsepower.
To convert watts to horsepower, we use the conversion factor \( 1 \, \text{HP} = 746 \, \text{W} \):
\[ \text{Power in HP} = \frac{-8000 \, \text{W}}{746 \, \text{W/HP}} \approx -10.72 \, \text{HP} \]

### Final Answer:
The power applied by the electric motor at full load is approximately \( -10.72 \, \text{HP} \).

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