Set 1 (Work and Energy)

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. You throw a zero point five kilogram ball up at twelve metres per second, and it comes back to your hand at only ten. Can you find the air resistance, and how high the ball went? Sounds impossible, right? By the end of this class, you will do it. This is set one of mixed JEE Main-pattern problems on work and energy: three quick ideas, then two full problems. So, let's begin.

0:31Idea 1: work by a constant force

Idea number one. The force F has a little arrow on top: it is a vector, four i hat plus three j hat, in newtons, capital N, okay? i hat and j hat, with the little caps, are unit vectors along x and y. It moves a body from the point one, minus one, to the point four, three, in metres. Find W, the work done. The body ends at four, three. So multiply F with four i hat plus three j hat: sixteen plus nine, twenty five joules, with capital J for joule, right? No, no, careful! Work needs the displacement, not the final position. For a constant force, W equals F dot s, the dot product, where s with its arrow is the displacement: final position minus initial position. With me? So x goes from one to four: three. And y goes from minus one to three: three minus, minus one, four. So s equals three i hat plus four j hat. Clear? Now the dot product: multiply the matching parts, and add. Four times three, plus three times four: twelve plus twelve, twenty four joules. Not twenty five!

1:51Idea 2: the work-energy theorem

Idea number two. The speed of a three kilogram body rises from two to six metres per second; m s to the power minus one means metres per second. Find the work done on it. The speed rises by four, so half times three times four squared: twenty four joules, right? Oh no, careful! Kinetic energy goes as the speed squared. The work-energy theorem: W net, with net written below, is the total work of all the forces. It equals delta K, the change in kinetic energy; delta, the little triangle, means change. So half m, times v squared minus u squared, with u the starting and v the final speed. Got it? Here: half times three, times thirty six minus four: forty eight joules. Double the trap answer! So, square first, then subtract.

2:43Idea 3: work on a spring from x₁ to x₂

Idea number three: springs. Here, a spring of constant k, two hundred and fifty newtons per metre, written N m to the power minus one, is compressed further from two to six centimetres; c m means centimetres. Find the work needed. It moves four centimetres more, so half k times zero point zero four squared: zero point two joules. Easy, isn't it? No, no, wait! Look at the figure. The stored energy, capital U, is half k x squared, and the restoring force F is minus k x. On the right, U rises faster and faster. The work needed from x one to x two, with small labels below x, is the rise in U between them: the jump in height of the curve, not the shaded area under it. Compressing works the same way. So it is half k, times x two squared minus x one squared, not the difference squared. Okay? In metres, x one is zero point zero two and x two is zero point zero six. Half k is one hundred and twenty five, and the squares are zero point zero zero three six and zero point zero zero zero four. With me? Subtract: zero point zero zero three two, times one hundred and twenty five, zero point four joules. Double the trap answer again! Same lesson: square first, then subtract.

4:12Example 1: pushing a crate up a rough incline

Now, example one. Find the work needed to push a twenty kilogram crate ten metres up a thirty degree incline, at constant speed, with a force along the incline. The small circle is the degree sign, and mu, the Greek letter, is the coefficient of friction, zero point one. Take g as ten metres per second squared. Pause the video, and try it first. Done? Okay, let's check. Constant speed, so no change in kinetic energy. So the work needed is zero, isn't it? No, no, careful! Zero is the total work, not the work of your push. In the figure, our crate is the block: your force F along the incline, marked pull, friction small f, the normal reaction capital N, and the weight m g; the dashed s is the displacement, and theta is the incline's angle. Each W carries its force as a small label below. The four works add to delta K, zero, and W N is zero, as the normal reaction is perpendicular to the motion. Clear? On the right, friction and gravity do negative work; your push must cancel them. W F equals m g sine thirty times s, plus mu m g cos thirty times s. Your push is along the incline, so the normal reaction is m g cos thirty, not m g. Got it? Numbers: m g is two hundred newtons, and s is ten metres. Sine thirty is zero point five, and mu cos thirty is zero point one times zero point eight six six. So two hundred, times the bracket, times ten: about one thousand one hundred and seventy three joules; the wavy equals sign means approximately. Not zero at all!

6:04Example 2: ball thrown up against air resistance

Now the harder one: our puzzle from the start. Here, a zero point five kilogram ball thrown up at twelve metres per second returns to the thrower at ten metres per second. Assuming a constant air resistance, small f, find f and the height reached. Pause and try. Hint: use the work-energy theorem twice. Done? Okay, let's check. Going up, the ball slows down and stops at the top. So which forces take away its kinetic energy? Gravity and the air, both, exactly! Both act downwards as it rises. m g is zero point five times ten, five newtons. So half times zero point five times twelve squared, thirty six joules, equals m g plus f, times h, the height reached. Okay? Now, coming down the same height h. The air acted downwards before, so it adds to gravity again, five plus f, yes? No, no, wait! Air resistance always opposes the motion. Now the ball moves down, so the air pushes up. So half times zero point five times ten squared, twenty five joules, equals m g minus f, times h. With me? Divide the two equations; h cancels. Thirty six times, five minus f, equals twenty five times, five plus f; the double arrow on screen just means, which gives. One hundred and eighty minus thirty six f equals one hundred and twenty five plus twenty five f. So sixty one f is fifty five. Clear? f is fifty five over sixty one, about zero point nine zero newtons. Then h equals thirty six over, five plus f, that is thirty six over five point nine zero: about six point one metres. Right? And now, our puzzle. Over the round trip, gravity's work is zero: what it takes going up, it gives back coming down. So the lost kinetic energy, thirty six minus twenty five, eleven joules, all went in working against the air: f times h going up, and f times h coming down, two f h. With our values, two f h is about eleven. Puzzle solved!

8:26Quick revision: dot product, work-energy theorem, springs

So, quick revision. Constant force: W equals F dot s, with the displacement. W net equals delta K, and spring work is half k times x two squared minus x one squared: square first, then subtract. Constant speed: the total work is zero, not the work of your push. And friction and air resistance always oppose the motion. Now try the other problems of this set on this page: solve each one yourself first, then check the solution. You can do this. All the best!