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## Homework Statement

Hello Everybody,

When i was searching for power required to lift an object, i found that:-

For Example, Q :- 100 kgs to be lifted 3 metres in 5 seconds. (vertical)

A :- Mass * Gravity * (Distance / Time)

= 100 * 9.8 * (3/5)

= 588 Watts

Assuming a efficiency loss of 22% (588/78%) :- 750 watt required to lift a object weighing 100 kgs, 3 metre high in 5 seconds

I don't think there is any problem in above calculation (pls correct if i'm wrong).... But my question is , What will be the

**monthly electricity consumption (KWH)**of 750 watt motor if run for the above said purpose

*5 times in*

**a minute (5 cycles)**(i.e run for 25 seconds/min)## Homework Equations

.

## The Attempt at a Solution

[/B]

A) Will the answer be :-

Total HOURS run in a DAY :- (25 seconds / 60 seconds) * 24 hrs a day

= 10 hrs

Total hrs in a month = 10 * 30 = 300 hrs in a month

Since 750 watt is 75% of 1 Kw = 300 * 75% =

**225 KWH**or

**225 units**( I think this makes more sense)

**Or**

B)

Total cycles in a month = 5 times in a minute * 60 min * 24 hrs * 30 days

= 216000 cycles

Power consumption per cycle = 750 watt

Hence total power consumption is = 750 watt * 216000 cycles

= 162000000 watts

=

**162000 Kwh**( I think this answer is absurd)

(I know this is a very basic doubt... but pardon me, since i don't belong to Physics field and i couldn't find a clear cut answer) May be i was looking in the wrong place..:) Sorry