How High Can a 12V Battery Lift an 8.1 kg Mass with a 90% Efficient Motor?

AI Thread Summary
A fully charged 12V battery rated at 4 Ampere-hours can provide 172,800 Joules of energy when connected to a 90% efficient motor. After accounting for efficiency, the motor can perform 155,520 Joules of work. Using the conservation of energy principle, this energy can lift an 8.1 kg mass to a height of approximately 1959.18 meters. The calculations are correct, and the large height value is reasonable given the low mass. The approach using conservation of energy is appropriate for this problem.
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Homework Statement


A fully energized 12V battery is rated as "4 Ampere-hours." Suppose this battery is connected to a motor that is 90% efficient at converting electrical to mechanical power. How high could this battery-supplied motor lift a 8.1 kg mass?


Homework Equations


V = I*R
P = V*I
P = (I^2)*R

The Attempt at a Solution



An amp-hour is how many amps a battery can supply for one hour.
Using the power equation:
P = (12V)*(4A*hr) = 48 Watt*hr

Since one watt-hr = 3600 Joules,
(48 W*hr)*(3600) = 1728 Joules

The motor is 90% efficient so,
1728 * .9 = 1555.2 Joules

I think the above is right (hopefully), but I'm not sure how to use the 1555.2 Joules to find the maximum height that this motor can lift the mass. Any help would be greatly appreciated. Thanks!
 
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Power isn't in watt-hours, a watt hour is energy

Right idea still, 12V*4Ahr is 48 Watt*hr, but that itself equals 172800 joules

90% of that is straight up how much work the motor can do. Work...mass...looking for distance...
 
So, now at 90% of 172,800 J = 155,520 J

Using conservation of energy:
155,520 J = mgh
155,520 J = (8.1 kg)*(9.8 m/s^2)*(h)
h = 1959.18 m ?

This seems like a large value but reasonable because the mass is not that great.
 
Before I submit my answer, would you agree with using conservation of energy and the answer that I got?
 

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