1. The problem statement, all variables and given/known data 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? 2. Relevant equations V = I*R P = V*I P = (I^2)*R 3. 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!