Work Done by Springs and by Friction

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Stretching a spring ten centimetres from its natural length takes one joule. Now stretch it ten centimetres more. One joule again, right? Hmm, are you sure? Today: work done by springs, and by friction, which has its own surprise. Both come up often in JEE, so let's begin.

0:26Hooke's law: the spring pulls back

So, look at the figure. On the left, a spring joins a wall to a box. On top, the spring has its natural length, and the dashed line marks x equals zero. Below, the box is pulled out by a distance x. Which way does the spring pull the box? Back towards the wall, yes! That is the arrow marked k x. Hooke's law: F equals minus k x. The minus sign means the force opposes x: stretched, the spring pulls back; compressed, it pushes out. That is the restoring force in the figure. Okay? Now, careful with x. It is the extension or the compression from the natural length, not the full length of the spring. Right? And k, the spring constant, tells how stiff the spring is. Its unit is newton per metre, written N, m to the power minus one. Clear?

1:18Work done by the spring, and by you

Now, the spring's own work. Its force changes with x, so we integrate: W s, with a small s for spring, equals the integral, the tall S, from x one to x two of minus k x, d x. Here x one and x two, with small subscripts, are the start and end deformations. Okay? That gives W s equals half k, times x one squared minus x two squared. Start squared, minus end squared. Got it? Now tell me. You stretch the spring, so x two is bigger than x one. Is the spring's work positive or negative? Negative, exactly! The spring pulls back as the box moves out. You, pulling slowly, just balance the spring and do the opposite work: half k, times x two squared minus x one squared. On the right is the graph of capital U, the stored energy, half k x squared. This work is the rise in U, from x one to x two: the gain in height of the curve, not the shaded area under it. With me? And x always comes squared. So stretching or compressing a spring by the same x needs the same work. Clear?

2:28Springs in series and in parallel

Now, two springs joined end to end, in series. Is the pair stiffer or softer than each spring alone? Softer, yes! In series the tension is the same in both, and the stretches add. So the reciprocals add: one over k equals one over k one, plus one over k two, the two constants with small subscripts. Okay? In parallel, side by side, both stretch equally and the forces add: k equals k one plus k two. Don't mix these up! Now cut a spring into two equal halves. Is each half stiffer or softer? Stiffer! Any piece that is a fraction f of the original length has constant k over f. Half the spring: f is one half, so the constant is two k. Clear?

3:17Friction: is its work always negative?

Now, friction. On a fixed surface, kinetic friction is mu k N: mu k, the Greek letter mu with a small k, is the coefficient of kinetic friction, and capital N is the normal force. It opposes the sliding, so its work is minus mu k N, times the distance slid. Okay? So, friction always does negative work, right? Every single time? No, no, careful! If a body does not move on a fixed surface, there is no displacement, so static friction does zero work. Now, a box sits on the floor of a truck that speeds up, and moves with it without slipping. Which force pushes the box forward? Static friction, exactly! It points forward, and in the ground frame the box moves forward, so here static friction does positive work. Common trap! Got it?

4:05Topper's trick: two surfaces sliding on each other

Now a topper's trick. When two surfaces slide on each other, like a block on a moving plank, the two friction forces, one on each body, are equal and opposite. So their total work is zero, right? No, no, wait! The two bodies move through different distances. The total work is minus f, the friction force, times the relative sliding distance: how far one surface slides over the other. Okay? This energy appears as heat: the heat produced is f times the relative sliding distance. No need to track each body. Clear?

4:42Example 1: stretching a spring from 10 cm to 20 cm

Now, remember our puzzle? Let's finish it. Example one: a spring with k equals two hundred newtons per metre is stretched further, from ten to twenty centimetres beyond its natural length. Find the work needed. Pause the video, and try it first. Done? Okay, let's check. The first ten centimetres, zero point one metre: half times two hundred, times zero point one squared, gives one joule, as in our puzzle. So the next ten centimetres needs one joule again, right? No, no, careful! The spring is already stretched now, so it pulls back harder all the way. Your work, pulling slowly, is half k, times x two squared minus x one squared, with x one equal to zero point one metre and x two equal to zero point two metre. Always metres, okay? Half of two hundred is one hundred. Zero point two squared is zero point zero four, and zero point one squared is zero point zero one. So one hundred times, zero point zero four minus zero point zero one: three joules! Three joules, not one! Work grows as x squared, so the second ten centimetres needs three times the first. Puzzle solved!

5:57Example 2: friction work on block and plank, and the heat

Now a harder one, JEE level. Here, a long plank is kept moving at a constant three metres per second on a smooth floor. Then a two kilogram block is placed gently on it; mu, the friction coefficient between them, is zero point three. Until it stops slipping, find the work done by friction on the block, on the plank, and the heat produced. Take g as ten metres per second squared. Pause and try. Hint: find how long it slips. Done? Okay, let's check. Kinetic friction drags the block forward, as the plank slides under it. The normal force is m g, so f equals mu m g: zero point three times two times ten, six newtons. The block's acceleration, a, is six over two: three metres per second squared. With me? The block speeds up until its speed, v, matches the plank's three metres per second; then the slipping stops. So the time, t, is three over three: one second. Right? In that second, the block moves half times three times one squared: one point five metres; the plank moves three times one: three metres. Now, friction's work on the block: positive or negative? Positive! Friction pushes the block forward, and the block moves forward: six times one point five, plus nine joules. Kinetic friction, yes, but on a moving surface, not a fixed one. Clear? On the plank, friction acts backwards while the plank moves forward three metres. Minus six times three: minus eighteen joules. Got it? Total work: nine minus eighteen, minus nine joules. Those nine joules are lost as heat. Check with our topper's trick: the relative sliding is three minus one point five, one point five metres. Six times one point five: nine joules of heat. Same answer, in one line!

7:56Quick revision: spring work, series and parallel, friction

So, quick revision. The spring does half k, times x one squared minus x two squared; you, pulling slowly, do the opposite. In series the reciprocals add, in parallel the constants add, and a fraction f of a spring has constant k over f. Kinetic friction on a fixed surface does negative work, but friction can also do positive work, and the heat is f times the relative sliding distance. Now try the practice problems on this page. All the best!