An electric motor rated at 1.5 kW is used to lift a 900 kg load through 600 cm. The motor has an efficiency of 80%. Calculate: a) Potential energy gained by the load.

Let's break down the problems from the image.

First, regarding the potential energy calculations:

• The problem "An object of 10kg is raised through 8m calculate its potential energy" resulting in $784.8 \, \text{J}$ was solved in a previous interaction.
• The problem "An electric motor rated at 1.5kW is used to lift 300kg load through 600cm... Potential energy gained by the load" resulting in $17658 \, \text{J}$ was also solved in a previous interaction.

Now, let's address the new calculations and the remaining parts of the electric motor problem.

: Efficiency Calculation (from the top right of the image)

Given:

• Mechanical Advantage (M.A.) = 4
• Velocity Ratio (V.R.) = 6

Step 1: State the formula for efficiency.

$$\eta = \frac{\text{M.A.}}{\text{V.R.}} \times 100\%$$

Step 2: Substitute the given values and calculate the efficiency.

$$\eta = \frac{4}{6} \times 100\%$$ $$\eta = \frac{2}{3} \times 100\%$$ $$\eta \approx 66.67\%$$

The efficiency is $\boxed{66.67\%}$.

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: Electric Motor (Hospital Lift) Calculations

Given:

• Motor power rating (Input Power, $P_{in}$) = 1.5 kW = $1500 \, \text{W}$
• Mass of load, $m = 300 \, \text{kg}$
• Height, $h = 600 \, \text{cm} = 6 \, \text{m}$
• Efficiency, $\eta = 80\%$
• Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

First, we need to calculate the useful output work (potential energy gained by the load).

$$\text{Output Work} = PE = mgh$$ $$\text{Output Work} = (300 \, \text{kg}) \times (9.81 \, \text{m/s}^2) \times (6 \, \text{m})$$ $$\text{Output Work} = 17658 \, \text{J}$$

Now, let's solve the specific parts requested: