Work Using Components and Along Paths

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. You carry your school bag up to the second floor, once by the short staircase, and once by the long, winding ramp. Gravity pulls on the bag the whole way. So tell me, on which route does gravity do more work on the bag? The longer route, surely? Hold on to your answer. Today: work from components, work along different paths, and a rope over a pulley. In JEE, questions like these come up often. So, let's begin.

0:34Work from components: no angle needed

So, often the force comes as a vector, in pieces. F with a small arrow on top: the arrow tells you it is a vector. i hat, j hat and k hat, the letters wearing little caps, are unit vectors along x, y and z. And F x, F y and F z, with the small letters below, are the components of the force. One condition: this force is constant, same size and same direction throughout. Okay? Next, the displacement, s with an arrow. It equals r f minus r i: the final position vector, minus the initial position vector. Careful, the small i below the r means initial; it is not the unit vector i hat. Its components are delta x, delta y and delta z; capital delta, the little triangle, means change. Clear? Now, look. Nobody has told us the angle between these two vectors. So how do we find the work? The dot product, yes, exactly! W equals F dot s; the dot in the middle is the dot product. And it equals F x times delta x, plus F y times delta y, plus F z times delta z. Multiply the matching components, and add. No angle needed! Just keep every minus sign with its component. Got it?

1:53The angle from the dot product

Now, sometimes the angle is asked too. Remember, the dot product also equals F s cos theta. Theta, this Greek letter, is the angle between the force and the displacement. Right? So just turn it around. Cos theta equals F dot s, over F times s. On top, the dot product from the components; at the bottom, the two sizes multiplied. With me? And notice, F and s at the bottom have no arrows. So they are magnitudes: the lengths of the vectors, never negative. Okay?

2:28Constant force: the path does not matter

Now, back to our puzzle. For a constant force, W equals F dot, r f minus r i, in brackets. Look at it carefully. Does the path appear anywhere in this formula? No, it doesn't! Only the starting point and the end point. So a constant force does the same work along every path between the same two points. This double arrow on screen is read as implies. Okay? So, the stairs or the ramp? Near the earth, gravity on the bag is a constant force, so it does the same work on both routes. Only the change in height matters. Puzzle solved! And if the particle comes back to where it started? Then r f equals r i, the displacement is zero, so the work is zero: zero work round a closed path. But remember, all this is only for a constant force, clear?

3:22Topper's trick: add the forces first

Now, a topper's trick. When several constant forces act together, add the forces first: F one plus F two, and so on. Then take one dot product with s. W net, the total work, in a single line. Okay? And if the question asks for the work done by one force only? Then use that force alone, and ignore the others completely, right?

3:47A rope pulled over a pulley

Now, one last idea. You pull a string over a pulley, and the string pulls a block. The tension, capital T, has the same size all along. But as the block moves, the string turns, so the direction of the tension keeps changing. Our dot product needs one fixed force. So now what? Simple. Just watch the string! W equals T times the length of string pulled through. Your hand pulls the free end through some length, and a light string over a smooth pulley passes that same work on to whatever it pulls. Clear?

4:22Example 1: work and angle from F and s components

Now, example one. F equals three i hat plus four j hat newtons; the capital N on screen stands for newtons. And s equals six i hat minus two j hat metres, the small m. Find the work, and the angle between the force and the displacement. Pause the video and try it. Done? Okay, let's check. The i parts: three times six, eighteen. The j parts: four times two, eight. So eighteen plus eight, twenty six joules, right? No, no, careful! Look at s again. Its j component is minus two, not two. So W equals three times six, plus four times minus two. That one minus sign is the whole trap. Got it? So, eighteen minus eight: ten joules. The capital J is for joules. Positive work, so the force helps the motion. Okay? Now the angle. First, the magnitudes. F equals the square root of three squared plus four squared: five newtons. And s equals the square root of six squared plus minus two squared, that is, root forty metres. The tick shaped sign is the square root, and the small two up top means squared. Clear? So cos theta equals ten, over five times root forty. That is about zero point three one six. With me? So theta is about seventy one point six degrees. The wavy equals sign means approximately, and the small circle means degrees. Less than ninety degrees, which matches our positive work. That's it!

5:59Example 2: work by a rope pulled over a pulley

Now a harder one. Read it with me. There is a block on a smooth floor, tied to a light rope that goes over a small fixed pulley, three metres above the block's level. The free end is pulled with a constant force of fifty newtons. The block moves from four metres away from the pulley, measured horizontally, to directly below it. Find the work done by the rope on the block. Pause the video and try it. Hint: watch the rope. Done? Okay. The pull is fifty newtons, and the block moves four metres. So the work is fifty times four, two hundred joules, right? Oh no, careful! That is the trap. The tension is fifty newtons, yes, but it acts along the rope, slanting up towards the pulley. And as the block moves, the rope turns, so its direction keeps changing. Fifty times four would need the full fifty newtons along the floor the whole way. Not the case! So, watch the rope instead. Okay? At the start, how long is the rope from the pulley to the block? Three metres up, four metres across. Square root of four squared plus three squared: five metres. The three, four, five triangle! Right? At the end, the block is directly below the pulley, so the rope there is just three metres. So tell me, how much rope has been pulled through the pulley? Five minus three: two metres, yes! The fifty newton pull at the free end draws two metres of rope through. Clear? So W equals T times two, fifty times two: one hundred joules. The light rope passes this same work on to the block. So one hundred joules, not two hundred. Remember the conditions: a light rope, a smooth fixed pulley, and a constant pull at the free end. That's it!

7:51Quick revision: components, any path, the pulley rope

So, quick revision. One: for a constant force, multiply the matching components and add. Two: a constant force does the same work on every path between two points, and zero work round a closed path. Three: for a rope over a pulley, W equals T times the length of rope pulled through. Now go and try the practice problems on this page. You can do them, I promise. All the best!