What power is needed to lift a 49-kg person a vertical distance of 5.0 m in 20.0 s ?

Options

A. 12.5W
B. 60W
C. 25W
D. 120W

Answer

To find the power needed to lift a 49-kg person a vertical distance of 5.0 m in 20.0 s, we can follow these steps:

### Step 1: Calculate the Work Done
The work done (W) to lift an object is given by the formula:
\[ W = mgh \]
where:
- \( m \) = mass of the person (49 kg)
- \( g \) = acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))
- \( h \) = height (5.0 m)

Substituting the values:
\[ W = 49 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 5.0 \, \text{m} \]

### Step 2: Perform the Calculation
Calculating the work done:
\[ W = 49 \times 9.8 \times 5.0 \]
\[ W = 49 \times 49 \]
\[ W = 2401 \, \text{J} \]

### Step 3: Calculate the Power
Power (P) is defined as the work done per unit time:
\[ P = \frac{W}{t} \]
where:
- \( t \) = time taken (20.0 s)

Substituting the values:
\[ P = \frac{2401 \, \text{J}}{20.0 \, \text{s}} \]

### Step 4: Perform the Calculation for Power
Calculating the power:
\[ P = 120.05 \, \text{W} \]

### Final Answer
The power needed to lift the 49-kg person a vertical distance of 5.0 m in 20.0 s is approximately **120.05 watts**.

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