1. The problem statement, all variables and given/known data A standard D cell can supply 10mA at 1.5 volts for about 300 hours. An alkaline D cell can do about the same. Assume - 2 - that the chemistry of the batteries will produce the same amount of energy and will maintain the same EMF regardless of the current drawn from the battery. However, the internal resistance of a standard D cell is about 1 ohm and the internal resistance of an alkaline cell is about 0.1 ohm, so the amount of energy dissipated internally is different. Let us suppose that you have a multi-speed winch that is 50% efficient and that your mass is 60kg. (a) If the winch is set to super-slow speed how high can one of these batteries lift you before it is exhausted? (b) What is the fastest rate (in cm/sec) that each battery can lift you? (c) At this fastest lift rate, what is the maximum height that you achieve before the battery dies? How long does it take? 2. Relevant equations P=I*V=W/t U(gravity)=mgh 3. The attempt at a solution a). I said power dissipated here = 1.5V*10mA = 0.015 W. 50% efficiency means 0.0075 W are used by the winch. 0.0075 * (3600 s/hr * 300 hr) = 8100 J supplied by the battery. Weight = 60 kg * 10 m/s^2 = 600 N, so total distance traveled = 8100/600 = 13.5 meters. b.) For a standard D cell, - maximum current draw = 1.5V/(R_internal) = 1.5 A. Power dissipated = 1.5 V * 1.5 A = 2.25 W. At 50% efficiency 1.125 W are used by the winch. - Power = Energy / time. P=mg(delta(h)/delta(t)) 1.125 W = 60kg * 10m/s^2 * (h/t), so h/t = speed = 0.1875 cm/s. For an alkaline D cell, same as above but multiplied by a factor of 10 = 1.1875 cm/s. c.) Total energy supplied by the battery over its lifetime = 0.01A*1.5V*(300 hr * 3600 s/hr) = 16,200 J. Time = Energy/power. For a standard D cell, t = 16,200 J / (2.25 J/s) = 7,200 s. For an alkaline D cell, t = 16,200 / 22.5 = 720 s. Total distance traveled = speed * time = 0.1875 cm/s * 7200 s = 13.5 m total.