OpenStax College Physics, Chapter 20, Problem 64 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. We consider an electric car that has a 12.0 volt battery and has a mass of 750 kilograms and it has an efficiency of 95 percent in converting power from the battery into mechanical power of motion and in different scenarios—(a), (b) and (c)— we are going to figure out what current the battery needs to supply in order to meet the demands of the scenario. So in scenario (a), we are told that the car is going to accelerate from rest to a final speed of 25.0 meters per second in 1.00 minute and how much current is needed in that case? So efficiency is the power output divided by the power supplied input and we can solve for P in by multiplying both sides by P in divided by efficiency and the reason we are doing that is because P in is the current supplied by the battery multiplied by its voltage and so we are gonna substitute IV in place of P in down here and then we will divide both sides by the voltage to solve for the current. And P output is the mechanical energy provided to the car— its change in kinetic energy, in other words— divided by time So that's one-half mass times final speed squared minus one-half mv i squared and then we substitute all of this in for P out and we do that in red here and I have this multiply by 1 over efficiency because we are taking this P out and dividing it by efficiency but since P out is a fraction and I find it confusing to divide a fraction by a fraction I am instead multiplying by the reciprocal of the denominator which is to say multiply by 1 over efficiency. Okay! So divide both sides by V here, we did that substitution I mentioned for P in replaced with IV and dividing both sides by V, we end up with the current is one-half mv f squared because the initial speed is zero we are told— it's starting from rest. So current is one-half mv f squared divided by efficiency multiplied by voltage of the battery multiplied by the time it takes to accelerate. So we have one-half times 750 kilograms times 25.0 meters per second squared divided by 0.95—efficiency— times 12.0 volts from the battery times 60.0 seconds which is 343 amps. In part (b), the car is now going to climb a height of 200 meters in 2.00 minutes, which we convert into seconds by multiplying by 60 seconds per minute so that's 120 seconds at a constant speed of 25.0 meters per second and it's applying a force of 5.00 times 10 to the 2 newtons to counteract friction and air drag. So just as before, we start by saying power input is power output divided by efficiency well, I guess... that was the second step, wasn't it? But anyway... and then we have to replace power output with the terms that are relevant to the scenario so we have the rate at which it's gaining potential energy so here's the potential energy gained— mg times the height climbed— and divided by time to get the rate at which that energy is gained so that's the power plus this other power term which is the force multiplied by the constant speed and you might be more familiar with seeing something like force times distance over time— this might be a more familiar way of writing power— and we can replace distance with speed times time so I can write this as force times speed multiplied by time because that's what distance is divided by t and you can see that the t's cancel and we are left with F times v so F times v is one of the formulas we can use for power. Now to have a common denominator, let's multiply this top and bottom by t so we have mgh plus Fvt over t that's the power output, the power input is IV again and so we have IV equals mgh plus Fvt over t times 1 over efficiency. And then dividing both sides by V, we solve for current and so that's mgh times Fvt over voltage times time times efficiency. So 750 kilograms times 9.80 newtons per kilogram times 200 meters plus 500 newtons times 25.0 meters per second times 120 seconds all divided by 12.0 volts from the battery times 120 seconds times 0.950—efficiency— and that gives 2.17 times 10 to the 3 amps needed from the battery. In part (c), let's begin with the formula saying that current equals the power output divided by voltage times efficiency and we got that from saying that, you know, power input is power output over efficiency and power input is current times voltage and then dividing both sides by V. Okay... similar to what we have been doing in parts (a) and (b). Then we substitute force times speed in for P out because in scenario (c), the car is just moving at constant speed with a constant force against the air resistance and friction and so we substitute in Fv for P out so the current then is 500 newtons times 25.0 meters per second divided by 12.0 volts times 0.950—efficiency— and that is 1.10 times 10 to the 3 amps. And we expected the answer in part (c) to be less than the answer in part (b) because in part (b), the car is going at the same speed resisting the same amount of friction and air resistance but additionally it's going up, it's increasing its altitude and so we expect (b) to be a larger current and so this makes us believe the answer because part (c) should be smaller current than part (b).