Question

In order to keep a leaking ship from sinking, it is necessary to pump 12.0 lb of water each second from below deck up a height of 2.00 m and over the side. What is the minimum horse-
power motor that can be used to save the ship?

Answer 1

Answer:

P = 0.14 hp

Explanation:

The power required by the ship is given as:

$P = \frac{Work}{Time} = \frac{Potential\ Eenrgy}{t}\\\\P = \frac{mgh}{t}$

where,

P = Power = ?

m = mass to pump = (12 lb)(1 kg/2.20 lb) = 5.44 kg

g = acceleration due to gravity = 9.81 m/s²

h = height = 2 m

t = time = 1 s

Therefore,

$P = \frac{(5.44\ kg)(9.81\ m/s^2)(2\ m)}{1\ s}\\\\P = 106.8\ W$

Converting to horsepower (hp):

$P = (106.8\ W)(\frac{1\ hp}{746\ W})$

P = 0.14 hp