Question

in a factory it is desired to lift 2000kg of metal through a distance of 12m in 1minute .find the minimum horsepower of the engine to be used​

Answer 1

Answer:

Given:

Mass of the metal m = 2000 kg

height of the lift  h = 12 m

Time t = 1 min = 60 sec

To Find:

The minimum horsepower of the engine to be used​

Solution:

We know that,

Work done (W) against Gravity

$\boxed{\sf W=mgh}$

$\sf \implies W=2000 \times 9.8 \times 12$

$\sf \implies W=235200 J$

We know that,

Power (P)

$\boxed{\sf P=\dfrac{W}{t} }$

$\sf \implies P=\dfrac{235200}{60} \;W$

$\sf \implies P=3920\;W$

According to the question,

We are asked to find the minimum horsepower of the engine to be used​

Therefore,

$\sf \implies P=\dfrac{3920}{746}$

Since,

746 W = 1 hp

Hence,

For 3920 W hp is $\sf P$ hp

It can be expressed mathematically as,

$\sf 746\;W \implies 1\;hp$

$\sf 3920\;W \implies P \;hp$

So,

$\sf P= \dfrac{3920}{746} \;hp$

$\sf \implies P=5.25 \;hp$

Hence,

The minimum horsepower of the engine to be used​ is 5.25 hp