A 1500-kg elevator moves upward with constant speed through a vertical distance of 25 m. How much work was done by the tension in the elevator cable ?

Options

A. 990J
B. 8100J
C. 140,000J
D. 370,000J

Answer

To solve the problem of how much work was done by the tension in the elevator cable, we can follow these steps:

### Step 1: Identify the given data
- Mass of the elevator (m) = 1500 kg
- Distance moved upward (d) = 25 m
- Acceleration due to gravity (g) = 10 m/s² (or 9.8 m/s² if more precision is needed)

### Step 2: Calculate the weight of the elevator
The weight (W) of the elevator can be calculated using the formula:
\[ W = m \cdot g \]
Substituting the values:
\[ W = 1500 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 15000 \, \text{N} \]

### Step 3: Determine the tension in the cable
Since the elevator is moving upward at a constant speed, the tension (T) in the cable must equal the weight of the elevator:
\[ T = W = 15000 \, \text{N} \]

### Step 4: Calculate the work done by the tension
Work (W) done by the tension can be calculated using the formula:
\[ W = F \cdot d \cdot \cos(\theta) \]
Where:
- \( F \) is the force (tension in this case)
- \( d \) is the distance moved
- \( \theta \) is the angle between the force and the direction of motion. Since both the tension and the displacement are in the same direction (upward), \( \theta = 0° \) and \( \cos(0°) = 1 \).

Thus, the work done by the tension is:
\[ W = T \cdot d \cdot \cos(0°) \]
Substituting the values:
\[ W = 15000 \, \text{N} \cdot 25 \, \text{m} \cdot 1 \]
\[ W = 375000 \, \text{J} \]

### Final Answer
The work done by the tension in the elevator cable is **375000 Joules**.

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