The input power to an elctric motor is `200 KW`. Its efficiency is `80%`. It operates a crane of efficiency `90%`. If the crane is lifting a load of `3.6` tonnes, the velocity with which the load moves is

Options

A. `8 ms^(-1)`
B. `4 ms^(-1)`
C. `2 ms^(-1)`
D. `40 ms^(-1)`

Explanation

To solve the problem step by step, we will follow the given information about the electric motor, crane, and load.

### Step 1: Calculate the Output Power of the Electric Motor
Given:
- Input Power of the motor, \( P_{\text{input}} = 200 \, \text{kW} \)
- Efficiency of the motor, \( \eta_{\text{motor}} = 80\% = 0.8 \)

Using the formula for efficiency:
\[ \eta = \frac{P_{\text{output}}}{P_{\text{input}}} \]
We can rearrange this to find the output power of the motor:
\[ P_{\text{output, motor}} = \eta_{\text{motor}} \times P_{\text{input}} = 0.8 \times 200 \, \text{kW} = 160 \, \text{kW} \]

### Step 2: Calculate the Input Power for the Crane
The output power of the motor becomes the input power for the crane:
\[ P_{\text{input, crane}} = P_{\text{output, motor}} = 160 \, \text{kW} \]

### Step 3: Calculate the Output Power of the Crane
Given:
- Efficiency of the crane, \( \eta_{\text{crane}} = 90\% = 0.9 \)

Using the efficiency formula again:
\[ P_{\text{output, crane}} = \eta_{\text{crane}} \times P_{\text{input, crane}} = 0.9 \times 160 \, \text{kW} = 144 \, \text{kW} \]

### Step 4: Calculate the Force Required to Lift the Load
Given:
- Mass of the load, \( m = 3.6 \, \text{tonnes} = 3.6 \times 1000 \, \text{kg} = 3600 \, \text{kg} \)
- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \)

The force required to lift the load (weight) is given by:
\[ F = m \times g = 3600 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 35316 \, \text{N} \approx 36000 \, \text{N} \]

### Step 5: Relate Power, Force, and Velocity
The power output of the crane can also be expressed in terms of force and velocity:
\[ P = F \times v \]
Where:
- \( P = 144 \, \text{kW} = 144000 \, \text{W} \)
- \( F = 36000 \, \text{N} \)

Rearranging the formula to find the velocity \( v \):
\[ v = \frac{P}{F} = \frac{144000 \, \text{W}}{36000 \, \text{N}} = 4 \, \text{m/s} \]

### Final Answer
The velocity with which the load moves is \( 4 \, \text{m/s} \).

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