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  1. I'd like to learn how much energy I need to lift a 200 kilograms weight on normal earth conditions?

  2. For example how much electric power do we need?

I'm not a physicist and not a student and this is not my hw:) I just wondering.

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The amount of energy needed is proportional to the mass of the object AND also proportional to the height over which it is to be lifted. Without specifying the lifting height your question can not be answered. –  Johannes Oct 7 '12 at 14:39
I'm sorry I should be more spesific. Mass of the object is I think should be 20 kg, so weight can be calculated as 200 kg with 10 m/s g; and height over is 50 meters. Also there could be a speed variable for this cause. –  Serhat Koroglu Oct 7 '12 at 15:02
Energy is force times distance, that is, weight times height. Power is energy per unit time, so if you lift it slow, it doesn't take much power. If you lift it fast, it takes a lot of power. After you run the numbers, it's just a matter of converting units. –  Mike Dunlavey Feb 21 '13 at 16:45
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1 Answer

up vote 1 down vote accepted

In absence of a numerical value for that height, we shall call it $h$. Then, the energy necessary to lift $20 \textrm{ kg}$ at ‘normal earth conditions’ (namely $9.81 \frac{\textrm{N}}{\textrm{kg}}$ acceleration due to gravity), is given by

$$ E = m \times g \times h = 20 \textrm{ kg} \times 9.81 \frac{\textrm{N}}{\textrm{kg}} \times h $$

With your new height of fifty metres, we can then plug this into our equation and get

$$ E = 9810 J\quad.$$

This energy is independent of the speed at which you lift the mass (unless it has extra speed left over when reaching the height of $50 \textrm{ m}$).

Furthermore, we are able to directly compute the minimum force necessary to lift an item of $20 \textrm{ kg}$ against the earth’s gravitational field, namely $F_{\textrm{min}} = 196.2 \textrm{ N}$. A larger speed then requires a larger force to accelerate the mass to this speed.

Note that the above calculations assume Newtonian gravity, and, more importantly, a frictionless system. While the assumption of Newtonian gravity usually holds at the surface of the earth, I am looking forward to see a real world lifting device working without losses due to friction.

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protected by Qmechanic Feb 21 '13 at 15:58

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