Question

What amount of water in kg would be lifted by a motor having power 0.5 hp at the height of 5 m in 98 seconds?

Answer 1

Answer on Question#40535, Physics, Mechanics

What amount of water in kg would be lifted by a motor having power 0.5 hp at the height of 5 m in 98 seconds?

Solution:

The motor provides the force that lifts the mass up a height $h$ against the force of gravity. Horsepower is a unit of power, which is work performed per unit time, $P = W/t$. If you know the power and the amount of time, the work that's performed by the motor is given by:

$$W = P \cdot t$$

where $P =$ power and $t =$ time.

Since the water is lifted through a height of $h = 5 \, \text{m}$, work done is equal to the potential energy $PE = \text{mgh}$.

$$W = mgh$$

So,

$$Pt = mgh$$

Solve for mass (m):

$$m = \frac{Pt}{gh}$$

One electrical horsepower is equal to 746 watts, $g =$ the acceleration due to gravity at the surface of the Earth (9.8 m/s²)

Thus,

$$m = \frac{Pt}{gh} = \frac{0.5 \cdot 746 \cdot 98}{9.8 \cdot 5} = 746 \, \text{kg}$$

Answer. $m = 746 \, \text{kg}$