Question

2. Find the H.P. required by the engine to lift 125 kg weight up to 12.5 m in 5 minutes?​

Answer 1

Answer:

Time = 5m = 150 second

P = mgh/t

P = 125 x 9.8 x 12.5 / 150

P = 102.08

HP = 0.183

Answer 2

Answer:

The horse power(H.P) required by the engine is 0.684H.P

Step-by-step explanation:

Given that,

The mass required to be lifted by engine is 125kg.

$m = 125 \: kg$

The height upto which the mass is to be lifted is 12.5m

$= > h = 12.5m$

And the time in which this entire process of lifting the mass through a certain height by the engine is 5 minutes.Converting it in SI Units,we get

$t = 5 \times 60 = 300sec$

Now the work done by engine is nothing but its potential energy(P.E)

$W=P.E=mgh$

Substituting the known values we get

$W = 125 \times 9.81 \times 12.5 \\ = 15328.125J$

The power consumed is ratio of work to time.Hence,

$P=\frac{W}{t}$

Substituting the given values we have

$P=\frac{15328.125}{30} = 510.9375W$

To convert it into Horse power we have an conversion as below

$1 H.P=746W$

Hence our calculated power in units of horse power becomes

$\frac{510.9375}{746} W = 0.684H.P$

Therefore,The horse power(H.P) required by the engine is 0.684H.P

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