Systems of Bodies, Strings and Internal Work

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. You jump straight up. The floor pushes on your feet, right? But while it pushes, your feet do not move at all, so the floor does no work on you. Zero work, and yet you leave the ground with kinetic energy. So where does that energy come from? By the end of this lesson, you will know. The same idea makes pulley and string problems really quick, and those come up often in JEE, so let's begin.

0:30Work and energy for a whole system

So, a system is two or more bodies treated together. External forces come from outside it; internal forces are the ones its parts exert on each other, okay? On screen, W with the small label e x t is the external work, and W with the label i n t is the internal work. Their sum equals delta K total. Delta, the little triangle, means change, so this is the change in the total kinetic energy. Clear? Now think. Internal forces always come in pairs, equal and opposite, by Newton's third law. So their works also cancel, and W int is always zero, right? Half right. For momentum, the pairs really do cancel, so internal forces cannot change the total momentum. Got it? But for work, no, no, careful! The two forces are equal and opposite, but the two parts can move through different distances, so their works need not cancel. So internal forces can change the total kinetic energy. That is the trap. Okay? You see this in an explosion, a person jumping, friction between two parts, and a spring between two blocks. Take the spring: two blocks at rest on a smooth floor, pushed apart by it. Both gain kinetic energy, but they move in opposite directions, so the total momentum stays zero. Right? And now, our puzzle. When you jump, the floor does no work, because your feet do not move while it pushes. Your kinetic energy comes from the internal work of your own muscles. That's it!

2:07A light string: the tension drops out

Now, strings. Light means the string's mass can be ignored, and inextensible means it cannot stretch. Over a smooth pulley, such a string has the same tension, capital T, at both ends. Since it cannot stretch, when one end moves a distance s, so does the other end. With me? Now picture one block rising and one falling. On one, the string pulls along the motion; on the other, against it. So what is the total work of the tension? Look at the two ends. On one end, plus T times s. On the other end, minus T times s. Equal and opposite works. Clear? Yes, exactly! They add up to zero. So the tension drops out of the system's energy equation, and you do not need it to find the speed. Perfect! But here is a trap. Some students say tension does no work at all. Wrong! On each body alone, it does work: plus T s on one, and minus T s on the other. It cancels only for the pair. Got it?

3:11Connected bodies: the quick recipe

So, for bodies connected by a light, inextensible string over a smooth pulley, the recipe. The work of gravity, plus the work of friction, equals the change in the total kinetic energy. The tension has already dropped out. Okay? Now, one body falls through a height h, and the other rises through the same height. Gravity pulls both down. So, is its work on the rising body plus, or minus? Minus, correct! Call the falling mass m one and the rising mass m two; the one and the two are the small subscripts. On the falling body, gravity's work is plus m one g h: its mass, times g, the acceleration due to gravity, times the height h. The rising body moves up against gravity, so it gets minus m two g h. With unequal masses, these two works do not cancel. Right? And a block sliding on a level table moves sideways, while gravity acts straight down. So gravity does no work on that block at all. Clear? Last, the kinetic energy. The string cannot stretch, so both bodies move with the same speed v. So K total is one half, times the bracket m one plus m two, times v squared. Starting from rest, this is the whole change in kinetic energy. With me?

4:30Example 1: speed of 4 kg and 1 kg masses over a pulley

Now let's use all this. Masses of four kilograms and one kilogram, k g on screen, hang from a light string over a smooth pulley, and are released from rest. Find their speed when the four kilogram mass has fallen one point five metres, small m for metres, with g as ten. Pause the video, and try it first. Done? Okay, let's check. Both masses and the string form one system, so the tension drops out. The four kilogram mass falls one point five metres: four times ten times one point five, plus sixty joules, capital J for joules. Okay? Now the one kilogram mass rises the same one point five metres, because the string cannot stretch. It moves up against gravity: minus one times ten times one point five, minus fifteen joules. Right? So the total work is sixty minus fifteen: forty five joules. The tension's two works cancel, so it drops out. Clear? Now the kinetic energy. Forty five joules equals one half, times the mass, times v squared. Which mass goes in here: four kilograms, or five? Five, yes! Both masses move with the same speed, so we use the total mass. Forty five equals one half times five times v squared, and the double arrow means this gives v squared equals eighteen. Got it? So v is the square root of eighteen, the root sign on screen, approximately four point two four; the wavy equals sign means approximately. And m s to the minus one means metres per second. We never found the tension. That's it!

6:14Example 2: block on a rough table pulled by a hanging mass

Now a harder one, with friction. Block A, of two kilograms, sits on a table, tied by a string over a pulley at the edge to a hanging block B, of one kilogram. The coefficient of friction between block A and the table is zero point two five; mu is the Greek letter on screen. Find the speed after block B has fallen zero point six metres from rest, again with g as ten. Pause and try. Hint: which block presses on the table? Done? Okay, let's check. Gravity first. Block B falls zero point six metres: one times ten times zero point six, plus six joules. And block A slides on the level table, so gravity does zero work on it, remember? Now friction. For that, we need the normal force on block A, so let's find it. The two blocks together weigh three times ten, that is thirty newtons. So the normal force is thirty newtons, right? No, no, careful! Only block A presses on the table; block B hangs in the air. So the normal force on block A, the slanted capital N, is two times ten: twenty newtons. The upright N after the number is the unit, newtons. Okay? Block A also slides zero point six metres, as the string cannot stretch. Friction opposes its motion, so its work is minus zero point two five times twenty times zero point six: minus three joules. With me? Now add: six minus three, three joules. So friction took away half. That equals one half times three times v squared, with the total mass of three kilograms, since both blocks move together. So v squared equals two. Clear? So v is root two, approximately one point four one metres per second. The tension never came in, and only block A pressed on the table. That's it!

8:15Quick revision: system work, string tension, pulley recipe

So, quick revision. For a system, external plus internal work equals the change in total kinetic energy, and internal work can change the kinetic energy. For a light, inextensible string, the tension's works cancel for the pair. And for connected bodies, the work of gravity plus the work of friction equals the change in total kinetic energy. Now go and try the three practice problems on this page. All the best!