Non-conservative Forces: W_(nc)=ΔE

Transcript

0:00Introduction

Hello everyone! Okay, a small mystery to start. Let a two kilogram block slide from rest down a curved track, five metres high. From our last lesson, on a smooth track it would reach the bottom at ten metres per second, right? But this track is rough, and the block arrives at only eight. So tell me, what happened to the energy that went missing? Hold that thought. The title gives the answer: W with a small n c below it, the work done by non-conservative forces, equals delta E, the change in mechanical energy; delta, the little triangle, means change. Friction problems like this come up often in JEE, so let's build it properly.

0:45Split the work into two parts

So, let's split the forces into two teams. Conservative forces, like gravity and the spring force; and non-conservative forces, like friction and air drag. Now tell me: the total work done by all the forces together equals what? The change in kinetic energy. Yes, exactly! The work energy theorem. So, W sub c plus W sub n c equals delta K, where capital K is the kinetic energy. W is work; the small c below means conservative forces, and the small n c means non-conservative ones. Okay? Now, the conservative part. Earlier in this chapter we saw that the work done by a conservative force is minus the change in potential energy. So, W sub c equals minus delta U, where capital U is the potential energy. Right? Now put this into the first line: minus delta U, plus W sub n c, equals delta K, okay? So what is W sub n c? Delta K plus delta U! Correct! Just move the minus delta U across. And K plus U is the mechanical energy, E, with the short word mech written below it. So, W sub n c equals the change in mechanical energy. Write this down; this one line is the heart of today's lesson. Clear?

2:05Which way does mechanical energy go?

Now, the sign. Friction acts on a block sliding along a rough floor. Is its work positive, or negative? Negative. Yes! Here friction opposes the sliding, so it takes energy away. Friction and air drag make W sub n c less than zero; that is the less than sign. So the mechanical energy falls. Got it? And that lost mechanical energy becomes heat: the block and the floor get slightly warmer. Energy is not destroyed; it just leaves the mechanical account. With me? Now the other way round. An engine pulling a car, or a person pushing a box, can feed energy in. Then W sub n c is greater than zero, the greater than sign, and the mechanical energy rises. Okay? And if the non-conservative forces do no work at all, W sub n c is zero, so the mechanical energy is conserved. That is our last lesson, as a special case. Nice, right?

3:02Topper's trick: rough inclines and floors

Now, a trick that toppers use. Suppose every rough surface has the same mu; mu, this Greek letter, is the coefficient of friction. Straight rough inclines, flat ground, any mix of them. Clear? Take one incline at an angle theta, another Greek letter, with length L along the slope. How much work does friction do on a block sliding down it? On the incline, the normal force is m g cos theta, so friction is mu m g cos theta. It opposes the motion, so W sub f, the work done by friction, is minus mu m g cos theta times the length along the slope. But look: L cos theta is just the base of the incline, its horizontal run. Right? So on every piece, sloping or flat, the friction work is minus mu m g times its horizontal distance. Add up the pieces: minus mu m g times the total horizontal distance. The angles drop out completely! Okay? Now release a block from rest at height h, and let it slide until it stops. Our main line, W sub n c equals delta K plus delta U, as always. Call the total horizontal distance x. Then minus mu m g x equals zero, minus m g h. With me? Cancel minus m g from both sides. So x equals h over mu! No mass, no g, no angles. Only one condition: the block must not stick earlier on an incline. Got it?

4:33Example 1: work done by friction on a rough curved track

Now, remember our mystery block? Let's solve it. So, a two kilogram block slides down a rough curved track from a height of five metres, and reaches the bottom at eight metres per second. Find the work done by friction, with g equals ten. On screen, k g is kilograms, m is metres, and m s to the power minus one is metres per second. Pause the video and try it. Done? Okay, let's check. The block starts from rest, so delta K is half times two times eight squared, minus zero: sixty four joules; the capital J stands for joules. Okay? Now, the change in potential energy. m g h is two times ten times five, that is one hundred joules. So delta U is plus one hundred joules, right? No, no, careful! The block goes down, so its potential energy falls. Delta U equals minus m g h: minus one hundred joules. This sign slip is a very common mistake in these questions. Clear? Now the main line. Friction is the only non-conservative force doing work here, so W sub f, the work done by friction, equals delta K plus delta U: sixty four minus one hundred, minus thirty six joules. Negative, as friction should be. Mystery solved: thirty six joules of mechanical energy became heat, and that is why the block arrived at eight, not ten, metres per second. Got it?

6:04Example 2: how far a block slides after a rough incline

Now a harder one, where our trick shines. So, a block is released from rest at the top of a rough thirty degree incline of height one point five metres; the small circle on screen means degrees. It slides onto rough level ground, and mu is zero point two everywhere. How far does it go on the level ground? Pause the video and try it. Done? Okay, let's check. But first, the condition from our trick. Will the block even start to slide, or will it just stick at the top? It slides. Correct! The block slides if tan of the angle is greater than mu. Tan thirty degrees is about zero point five seven seven; the wavy equals sign means approximately. That is greater than zero point two, so yes, it slides. With me? Now the trick. Same mu everywhere, so the total horizontal distance is h over mu: one point five over zero point two, that is seven point five metres. No mass, and no g needed. Okay? So, the answer is seven point five metres on the level ground, right? No, no, wait! That is the trap. Seven point five metres is the total horizontal distance, and it includes the base of the incline. The base is one point five times cot thirty degrees; cot is the cotangent, and cot thirty degrees is root three, that square root sign. So the base is one point five root three, about two point six zero metres. Clear? So, on the level ground: seven point five minus two point six zero, about four point nine metres. See? The whole incline, slope and all, took just one line. Got it?

7:49Quick revision: Wnc = change in E, signs, h/mu shortcut

So, quick revision. One: W sub n c equals delta K plus delta U, the change in mechanical energy. Two: friction and drag make it negative; an engine or a person makes it positive. Three: with the same mu everywhere, a block released from rest at height h travels a total horizontal distance of h over mu. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!