A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 54 m in 3.0 min, starting and ending at rest. The elevator's counterweight has a mass of only 950 kg, and so the elevator motor must help. What average power is required of the force the motor exerts on the cab via the cable ?

Answer

To solve the problem of determining the average power required by the elevator motor, we can follow these steps:

### Step 1: Convert Time to Seconds
The time given for the elevator to travel upward is 3.0 minutes. We need to convert this time into seconds for our calculations.

\[ \text{Time in seconds} = 3.0 \, \text{min} \times 60 \, \text{s/min} = 180 \, \text{s} \]

**Hint:** Always convert time into seconds when working with power calculations in physics.

### Step 2: Calculate the Weight of the Elevator and Counterweight
The weight of the elevator cab can be calculated using the formula:

\[ \text{Weight of the elevator} (W_e) = m_e \cdot g \]
where \( m_e = 1200 \, \text{kg} \) and \( g = 9.81 \, \text{m/s}^2 \).

\[ W_e = 1200 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 11772 \, \text{N} \]

The weight of the counterweight can be calculated similarly:

\[ \text{Weight of the counterweight} (W_c) = m_c \cdot g \]
where \( m_c = 950 \, \text{kg} \).

\[ W_c = 950 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 9319.5 \, \text{N} \]

**Hint:** Remember to use the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)) when calculating weight.

### Step 3: Determine the Net Force Acting on the Elevator
The net force that the motor must overcome is the difference between the weight of the elevator and the weight of the counterweight:

\[ F_{\text{net}} = W_e - W_c = 11772 \, \text{N} - 9319.5 \, \text{N} = 2442.5 \, \text{N} \]

**Hint:** The motor needs to exert enough force to not only lift the elevator but also to counteract the weight of the counterweight.

### Step 4: Calculate the Work Done by the Motor
The work done by the motor can be calculated using the formula:

\[ \text{Work} (W) = F_{\text{net}} \cdot d \]
where \( d = 54 \, \text{m} \).

\[ W = 2442.5 \, \text{N} \times 54 \, \text{m} = 132,045 \, \text{J} \]

**Hint:** Work is the product of force and displacement in the direction of the force.

### Step 5: Calculate the Average Power Required
Power is defined as the work done per unit time:

\[ \text{Power} (P) = \frac{W}{t} \]

Substituting the values we have:

\[ P = \frac{132,045 \, \text{J}}{180 \, \text{s}} \approx 733.5 \, \text{W} \]

**Hint:** Power is a measure of how quickly work is done, so always divide the total work by the time taken.

### Final Answer
The average power required of the force the motor exerts on the cab via the cable is approximately **733.5 W**.