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

$100\ \mathrm{kg}$ to be lifted 3 metres in 5 seconds. (vertical)


$$\begin{align} \text{mass}\times\text{gravity}\times\frac{\text{distance}}{\text{time}} &= 100 \times 9.8 \times \frac{3}{5} \\ &= 588\ \text{watts} \end{align}$$ Assuming an efficiency loss of 22% (588/78%), this gives 750 watts required to lift a object weighing $100\ \mathrm{kg}$ 3 metres high in 5 seconds.

I don't think there is any problem in the above calculation. But then I want to calculate the monthly electricity consumption (kWh) of a 750 watt motor if run for the above said purpose 5 times per minute, i.e run for 25 seconds each minute at full capacity of 750 watt.

I have two ideas:

  1. Is it $$\begin{align} \text{Total cycles in a month} &= \frac{\text{5 cycles}}{\mathrm{min}}\times\frac{60\ \mathrm{min}}{\mathrm{hr}}\times\frac{24\ \mathrm{hr}}{\text{day}}\times 30\ \text{day} \\ &= 216000\ \text{cycles} \end{align}$$ With power consumption per cycle of 750 watts, the total power consumption is $$750\ \mathrm{W}\times 216000\ \text{cycles} = 162000000\ \mathrm{W} = 162000\ \mathrm{kWh}$$ I think this answer is absurd

  2. Or $$\begin{align} \text{Total hours run in a day} &= \frac{25\ \text{seconds}}{60\ \text{seconds}}\times\frac{24\ \mathrm{hr}}{\text{day}} \\ &= 10\ \mathrm{hr} \\ \text{Total hours in a month} &= 10 \times 30 = 300\ \text{hours in a month} \end{align}$$ Since $750\ \mathrm{W}$ is 75% of $1\ \mathrm{kW} = 300 \times 75\% = 225\ \mathrm{kWh}$ or 225 units (I think this makes more sense)

  • $\begingroup$ Simply calculate the total amount of work that needs to be performed: number of lift cycles times the energy consumed per lift. $\endgroup$
    – CuriousOne
    Commented Oct 26, 2015 at 2:52
  • $\begingroup$ I get it now,Curious.. In 100 kg to be lifted 3 metres in 5 seconds = 750 Watts (pls reply whether this is right). So 750 watt if run for 1 hour it will consume 0.750 KWH. I thought 750 watt is for 5 seconds $\endgroup$
    – user96242
    Commented Oct 26, 2015 at 22:42

2 Answers 2


In your first option, the $750\ \mathrm{W}$ is already a rate. A watt quantifies energy per unit time. It doesn't make sense to multiply that by the number of cycles; you need to multiply it by a time to get an amount of energy. (As in $\mathrm{kWh}$, kilowatt-hour, which is a kilowatt times an hour.)

By analogy, it would be like saying your car goes 60 miles per hour, and you drive it for 8 hours in a day, then at the end of the day your car has traveled 480 mph. Do you see why that doesn't make sense? That's nearly the same mistake you made in your first approach.

There are a couple ways you can calculate the amount of energy needed: either find the total time taken by all those operational cycles and multiply that by the power of $750\ \mathrm{W}$, or as CuriousOne mentioned in a comment, find the amount of energy taken per cycle and multiply that by the number of cycles. I highly encourage you to track the units through your calculation; they can be multiplied and divided just like variables. If you had been paying attention to the units, you would have immediately noticed something was wrong when you came out with a number in watts ($162000000\ \mathrm{W}$, or if you were being more careful it would have been $1.62\times 10^8\ \mathrm{W}\,\text{cycles}$) despite needing a result in kilowatt-hours. You can't arbitrarily change units of power like the watt to units of energy like the kilowatt-hour.

  • $\begingroup$ i.e. $1\,\text{W} \times 1\,\text{cycles}\neq 1\,\text{Wh}$ $\endgroup$
    – innisfree
    Commented Oct 26, 2015 at 6:59
  • $\begingroup$ @innisfree yeah, I added a bit more to emphasize that. $\endgroup$
    – David Z
    Commented Oct 26, 2015 at 7:11
  • $\begingroup$ @innisfree Thanks David & innisfree.. So in the above question i have 216000 cycles of 5 seconds each.. i.e (216000 cycles*5 seconds)/3600 seconds = 300 hours of operation at full capacity...Which means 300 hours * (750 watt/1000 watt) = 225 Kwh or 225 units right? kindly reply whether this is right.. And thks for that miles hour example .. $\endgroup$
    – user96242
    Commented Oct 26, 2015 at 22:53
  • $\begingroup$ @user96242 300 hours * 750 W / 1000 W gives you a time, not an energy. $\endgroup$
    – David Z
    Commented Oct 27, 2015 at 4:40
  • $\begingroup$ Thanks David.. I've posted an answer in the thread. I think i've got it right now. Kindly reply. $\endgroup$
    – user96242
    Commented Oct 27, 2015 at 22:41

1) As per my question Total energy consumed (Kwh) will be the same IRRESPECTIVE of Power (Kw)

2) What matters as per my question is the "work done" (Energy consumed in joules)

3) Power is derived from the rate of energy consumed.. ( Here, in how much seconds i want a cycle to be completed)

So total energy consumed = [(100 kg * 9.8 * 3 metres) / 78% efficiency] * 216000 cycles = 810000000 Joules= 810000000 watt seconds

Since i want the power consumed in kwh = 810000000 watt seconds /3600 seconds = 225000 Watts Hour= 225 Kwh = 225 Units of ENERGY CONSUMED

Hence to derive the power = Work done per cycle / time required per cycle = 3750 joules / 5 sec = 750 Watts of POWER Required

If i want it to be completed in 4 seconds = Power required will be 937.5 Watts (3750 Joules/ 4 seconds)

But the total ELECTRICITY CONSUMPTION to complete the said work will be 225 units or KWH, irrespective of whether the cycle is 4 or 5 seconds each...


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