Average and Instantaneous Power

Transcript

0:00Introduction

Hello everyone! Let's start with two friends and one staircase. Same mass, same stairs. One walks up slowly. The other runs up. Both lift the same mass through the same height, so both do the same work against gravity. And yet, everyone says the runner is more powerful. So if the work is the same, what is different? Keep your answer in mind; we will settle it soon. Power comes up often in JEE, so let's build it properly. Let's go.

0:31Power: the rate of doing work

So, back to our two friends. Same work against gravity, okay? Then what is different for the runner? The time! Yes, exactly! The runner does the same work in less time. And that is what power measures: the rate of doing work, how fast the work gets done. With me? So, the average power over a time t is P sub av equals W over t: the total work done, divided by the time taken. Capital P is power, and the small letters a v written below it are a subscript, short for average. Clear? But work is not always done at a steady rate. So we also need the power at one instant: the instantaneous power. P equals d W over d t, the derivative of the work with respect to time. Okay?

1:21Instantaneous power: P = F dot v

Now, a much quicker way to get it. Work is force dot displacement, remember? So its rate is force dot velocity. P equals F vector dot v vector, which equals F v cos theta. The arrows on top mean vectors, the dot in the middle is the dot product, and theta, the Greek letter, is an angle. Clear? Which angle? Theta is the angle between the force and the velocity, at that instant. So if the force acts along the velocity, cos theta is one, and the power is simply F times v. Right? Now, gravity on a falling body. The force m g is straight down, so only the vertical part of the velocity counts. P equals m g times v sub y, the downward vertical component of the velocity; the small y written below v is a subscript. On the way up, gravity opposes the motion, so its power is negative. Got it? Now tell me. Suppose a ball is thrown at an angle. At the very top of its path, what is the power of gravity on it? Zero! Correct! At the top, the velocity is horizontal, so its vertical part, v sub y, is zero. The force is down, the velocity is sideways: theta is ninety degrees, and cos ninety is zero. So at that instant the power of gravity is zero, even though gravity is still pulling. Interesting, right?

2:47Units: the watt, and a trap

Now, units. The S I unit of power is the watt. One watt equals one joule per second: capital W is the watt, capital J is the joule, and s to the power minus one means per second. Careful: here capital W is the watt, not work. Okay? You will also see the horsepower, written h p. One horsepower equals seven hundred and forty six watts. Got it? Now, a quick question. Your electricity bill counts kilowatt hours. Kilo, watt: so the kilowatt hour must be a unit of power, right? No, no, careful! This is a classic trap. One kilowatt is a power, and one hour is a time. Power times time is energy. One thousand watts for three thousand six hundred seconds: three point six times ten to the power six joules. So the kilowatt hour is a unit of energy, not power. Clear?

3:47Two useful results

Alright, two useful results. First: the area under the P t graph, power against time, is the work done. W equals the integral of P d t; the tall S shaped sign is the integral sign. Makes sense: power times time is work. With me? Second: a constant force acts on a body that starts from rest. Constant force means constant acceleration, a, so the velocity is v equals a t. Okay? So the power is F v, which is F times a t. Both F and a are constant, so the power grows in proportion to time. This curly sign, like a stretched alpha, means proportional to. Right? Now think. The power starts at zero and grows steadily up to its final value. So what is the average power over that time? Half the final power! Yes, exactly! Picture the P t graph: a straight line from zero, so the area under it is a triangle, half of t times the final power. Divide that work by t, and you get half the final power. But careful: only for a constant force, starting from rest. Got it?

5:01Example 1: power of a boy running up stairs, in W and hp

Now, remember our runner from the start? Let's find his power. So, a sixty kilogram boy runs up thirty steps, each twenty centimetres high, in twelve seconds. Find his power in watts and in horsepower. Take g equals ten. Pause the video and try it. Done? Okay, let's check. First, the height. Thirty steps, twenty each: thirty times twenty is six hundred. So the height is six hundred metres, right? Oh no, careful! Those were centimetres! Six hundred centimetres is just six metres. Always change to S I units before you calculate. Okay? Now the work he does against gravity: W equals m g h. Sixty times ten times six: three thousand six hundred joules. With me? So his average power is W over t: three thousand six hundred over twelve, that is three hundred watts. And his friend who walked? Same work, but much more time, so less power. Puzzle solved! Right? Finally, horsepower. One horsepower is seven hundred and forty six watts, so we take three hundred over seven hundred and forty six; this wavy equals sign means approximately. About zero point four horsepower. Nice and quick!

6:28Example 2: power at t = 3 s and average power from rest

Now a slightly harder one. So, a constant force of ten newtons, the capital N on screen, acts on a two kilogram body starting from rest on a smooth floor. Find the power at t equals three seconds, and the average power over the first three seconds. Pause the video and try it. Done? Okay, let's check. Smooth floor, so no friction: the only horizontal force is our ten newtons. So a equals F over m: ten over two, five metres per second squared. Okay? It starts from rest, so v equals a t: five times three, fifteen metres per second. Right? The force acts along the velocity, so cos theta is one, and P equals F v. Ten times fifteen: one hundred and fifty watts. That is the power at three seconds. Clear? Now the average power over the first three seconds. That is just one hundred and fifty watts again, right? No, no, wait! One hundred and fifty watts is the power at the end. For the average, we need the total work. By the work energy theorem, the work done equals the gain in K E, half m v squared. Half times two times fifteen squared: two hundred and twenty five joules. With me? So P sub av is two hundred and twenty five over three: seventy five watts. And look: seventy five is exactly half of one hundred and fifty! Constant force, from rest: average is half the final power. That's it!

8:07Quick revision: P = Fv cos theta, kWh, half-power rule

So, quick revision. One: average power is W over t, and instantaneous power is F vector dot v vector, F v cos theta. Two: work is the area under the P t graph, and the kilowatt hour is a unit of energy, not power. Three: for a constant force from rest, the average power is half the final power. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!