(a)

Find the useful power output (in W) of an elevator motor that lifts a 2500kg load a height of 25.0m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10,000 kg—so that only 2500kg is raised in height, but the full 10,000 kg is accelerated. (Enter a number.)
W

(b)

What does it cost (in cents), if electricity is $0.0900 per kW · h? (Enter a number.) 

### Part (a): Finding the Useful Power Output

To find the useful power output, we need to calculate the work done by the elevator motor and the time over which this work is done. The work consists of two components:
1. **Work done against gravity** to lift the 2500 kg load.
2. **Work done to accelerate the entire 10,000 kg system** from rest to 4.00 m/s.

#### 1. Work done against gravity:

The work done to lift the load is given by:
\[
W_{\text{gravity}} = mgh
\]
where:
- $ m = 2500 \, \text{kg} $ (mass of the load)
- $ g = 9.81 \, \text{m/s}^2 $ (acceleration due to gravity)
- $ h = 25.0 \, \text{m} $ (height)

So,
\[
W_{\text{gravity}} = 2500 \times 9.81 \times 25.0 = 612,625 \, \text{J}
\]

#### 2. Work done to accelerate the system:

The work done to accelerate the mass is given by:
\[
W_{\text{kinetic}} = \frac{1}{2}mv^2
\]
where:
- $ m = 10,000 \, \text{kg} $ (total mass being accelerated)
- $ v = 4.00 \, \text{m/s} $ (final velocity)

So,
\[
W_{\text{kinetic}} = \frac{1}{2} \times 10000 \times (4.00)^2 = 80,000 \, \text{J}
\]

#### 3. Total work done:

The total work done by the motor is the sum of the work against gravity and the work to accelerate the system:
\[
W_{\text{total}} = W_{\text{gravity}} + W_{\text{kinetic}} = 612,625 + 80,000 = 692,625 \, \text{J}
\]

#### 4. Power output:

Power is the rate of doing work, so:
\[
P = \frac{W_{\text{total}}}{t}
\]
where $ t = 12.0 \, \text{s} $.

\[
P = \frac{692,625}{12.0} = 57,718.75 \, \text{W}
\]

So, the useful power output is **57,718.75 W**.

### Part (b): Cost of Electricity

Given:
- Power output = $ 57,718.75 \, \text{W} $
- Cost per kWh = \$0.0900

First, convert the power to kilowatts:
\[
P = 57.71875 \, \text{kW}
\]

Assuming the power is used for 1 hour, the cost is:
\[
\text{Cost in dollars} = P \times 1 \, \text{h} \times 0.0900 \, \text{\$/kWh} = 57.71875 \times 0.0900 = 5.1946875 \, \text{\$}
\]

Convert to cents:
\[
\text{Cost in cents} = 5.1946875 \times 100 = 519.46875 \, \text{cents}
\]

So, the cost is **519.47 cents** (rounded to two decimal places).

Would you like more details or have any questions?

Here are 8 related questions:
1. How would the power output change if the height lifted was 50.0 meters?
2. What would be the power output if the load was increased to 3000 kg?
3. How does increasing the time to 20.0 seconds affect the power output?
4. What is the cost of electricity if the motor runs for 2 hours instead of 1?
5. How would the calculation change if the velocity increased to 6.00 m/s?
6. What is the significance of the counterbalanced mass in this scenario?
7. How would the power output be affected if the motor had 90% efficiency?
8. What is the energy consumption in kWh for a 24-hour operation?

**Tip:** Always ensure to account for all forces acting on a system (e.g., gravity and acceleration) when calculating power and work.

Learn how to calculate the useful power output of an elevator motor lifting a load and accelerating a system, along with the cost of electricity consumed. Includes step-by-step calculations and formulas for work, energy, and power.

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