If a cage containing a truck of coal weighing 750 kg is raised to a height of 90 m in a minute, what is the total power expended? \( \left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}.\right) \)

To calculate the total power expended, we first need to determine the work done against gravity to lift the coal truck. The work done (W) is given by the formula: \[ W = m \cdot g \cdot h \] where: - \( m = 750 \, \text{kg} \) (mass of the truck) - \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity) - \( h = 90 \, \text{m} \) (height) Plugging in the values: \[ W = 750 \, \text{kg} \cdot 10 \, \text{m/s}^2 \cdot 90 \, \text{m} = 675000 \, \text{J} \] Now, power (P) is defined as work done per unit time. The time taken is 1 minute, which is 60 seconds. Thus, the power expended is: \[ P = \frac{W}{t} = \frac{675000 \, \text{J}}{60 \, \text{s}} = 11250 \, \text{W} \] So, the total power expended is 11250 watts or 11.25 kilowatts.