An escalator is 30.0 meters long and slants at `30.0^(@)` relative to the horizontal. If it moves at 1.00 m/s, at what rate does it do work in lifting a 50.0-kg woman from the bottom to the top of the escalator ?

Options

A. 49.3W
B. 245W
C. 98.0W
D. 292W

Answer

To solve the problem of how much power the escalator exerts while lifting a 50.0 kg woman, we can follow these steps:

### Step 1: Identify the known values
- Length of the escalator (L) = 30.0 m
- Angle with the horizontal (θ) = 30.0°
- Velocity of the escalator (v) = 1.00 m/s
- Mass of the woman (m) = 50.0 kg
- Acceleration due to gravity (g) = 9.8 m/s²

### Step 2: Calculate the height the woman is lifted
The height (h) can be calculated using the sine function:
\[ h = L \cdot \sin(\theta) \]
Substituting the known values:
\[ h = 30.0 \cdot \sin(30.0°) = 30.0 \cdot 0.5 = 15.0 \text{ m} \]

### Step 3: Calculate the weight of the woman
The weight (W) of the woman can be calculated using the formula:
\[ W = m \cdot g \]
Substituting the known values:
\[ W = 50.0 \cdot 9.8 = 490 \text{ N} \]

### Step 4: Calculate the work done (W_d) in lifting the woman
The work done against gravity is given by:
\[ W_d = W \cdot h \]
Substituting the known values:
\[ W_d = 490 \cdot 15.0 = 7350 \text{ J} \]

### Step 5: Calculate the time taken (t) to reach the top of the escalator
The time taken can be calculated using the formula:
\[ t = \frac{L}{v} \]
Substituting the known values:
\[ t = \frac{30.0}{1.0} = 30.0 \text{ s} \]

### Step 6: Calculate the power (P) exerted by the escalator
Power is defined as the work done per unit time:
\[ P = \frac{W_d}{t} \]
Substituting the known values:
\[ P = \frac{7350}{30.0} = 245 \text{ W} \]

### Final Answer
The rate at which the escalator does work in lifting the woman is **245 Watts**.

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