Vehicles, Lifts and Pumps

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Here is a car whose engine gives thirty kilowatts. On a level road, flat out, its greatest speed is twenty five metres per second. Now the road starts to climb, just gently. The driver keeps the engine at full power, the same thirty kilowatts. And every driver knows what happens: the car slows down on the hill. But wait. The power did not change! So where did the speed go? Think about it. Today we will find out exactly, and we will use the same idea for lifts and for water pumps too. Questions like these come up often in JEE, so let's begin.

0:42Climbing at a steady speed

So, here is our car, climbing a slope. Theta, the Greek letter that looks like an O with a line across it, is the angle of the slope. The orange arrow is F, the driving force from the engine, pointing up the slope. Two forces act the other way, down the slope. Capital R is the resistance to motion. And m g sine theta is the part of the car's weight along the slope, where m is the mass and g is the acceleration due to gravity. With me? Now, the car moves up at a steady speed, v. No speeding up, no slowing down. So tell me: how big must the engine's force F be? Yes, exactly! Steady speed means zero acceleration, so the net force is zero. F must balance both forces down the slope: F equals m g sine theta, plus R, okay? Now, remember from the last lesson: power equals force times velocity, so P equals F times v. So the power needed is the bracket, m g sine theta plus R, times v. The same two lines are written on the right of the figure too. And note the condition: this is for a steady speed. Clear?

1:56Full power: the greatest speed

Alright, now let's turn it around. Every engine has a greatest power; we call it capital P, the full power. At full power, P equals F times v, so the driving force F equals P over v, okay? So tell me. As the car goes faster and faster, what happens to the driving force? It gets smaller, yes! Bigger v, smaller P over v. So the car speeds up only until the driving force just equals the total resisting force. After that, the net force is zero, and the speed stops growing. That is the greatest speed: v max, the v with a small max written below it, equals P over the total resisting force. Got it? On a level road, that total is just R, the resistance. But up a slope, the total resisting force is m g sine theta plus R, both together. Same P on top, a bigger number below, so a smaller top speed. And that is our puzzle from the start, solved! The power is the same, but now gravity is also pulling the car back. Right? Now, below the top speed, the driving force is more than the total resisting force, so the car accelerates. Small a, the acceleration, equals one over m, times the bracket: P over v, minus R, minus m g sine theta. Driving force, minus everything that holds the car back, all over the mass. Just remember, v here is the speed the car has right now, not its top speed.

3:29Lifts and pumps

Clear? Now, lifts. Here the motor pulls the lift straight up at a steady speed v. Capital M is the total mass: the lift plus everyone inside it. Small f is the friction. So the motor must hold up the whole weight, capital M g, plus the friction: P equals the bracket M g plus f, times v. One common slip: students take only the mass of the lift and forget the passengers. Total mass, always! Okay? Next, pumps. Here m dot, the m with a small dot on top, is the mass rate: the kilograms of water the pump moves each second. Small h is the height the water is raised through, and v is the speed at which it is delivered, leaving the pipe. With me? So think. Each second, what energy does the pump give to the water? Two kinds! Potential energy, m dot times g h, because the water is raised. And kinetic energy, half m dot v squared, because it comes out moving. So P equals m dot, times the bracket g h plus half v squared. Oh no, careful here! When the water leaves with a speed, many students write only m dot g h and forget the half v squared. Don't do that! Right? Last point: no pump is perfect. This P is the useful power. The motor must put in more, so the input power equals the useful power over the efficiency. Divide by the efficiency, and the input always comes out bigger. Got it?

5:03Example 1: resistance from a 30 kW car's top speed

Now, Example one. Here, a car engine of power thirty kilowatts, k W on screen, gives a greatest speed of twenty five metres per second on a level road. The m s to the minus one means metres per second. Find the resistance. Pause the video and try it. Done? Okay, let's check. Greatest speed means the car has stopped speeding up. So the driving force exactly equals the resistance, capital R, okay? On a level road, the resistance is the only force holding the car back. So R equals P over v. And the power is thirty, so we put in thirty, right? No, no, wait! That thirty is in kilowatts. In the formula, power must be in watts, written capital W, so convert first. Thirty kilowatts equals thirty thousand watts. Clear? Now R equals thirty thousand over twenty five, which is twelve hundred newtons; capital N stands for newtons. So the resistance is twelve hundred newtons. See how quick that was?

6:10Example 2: greatest speed of the car up a slope

Now a harder one, Example two, and it is our puzzle from the start. The same car, of mass one thousand kilograms, k g on screen, climbs a slope where sine theta is point zero five. The resistance is the same as before. Find its greatest speed, taking g as ten. Pause the video and try it. Hint: what holds the car back? Done? Okay, let's check. First, the pull of gravity along the slope. m g sine theta equals one thousand times ten times point zero five, that is five hundred newtons. Right? Now, the engine gives thirty thousand watts, and gravity holds the car back with five hundred newtons. So v max equals thirty thousand over five hundred, sixty metres per second, correct? Oh no, careful! Sixty metres per second, faster than on the level road? That cannot be right. We forgot the resistance. It does not vanish on a slope! So the total resisting force is twelve hundred plus five hundred: seventeen hundred newtons. With me? Now v max equals thirty thousand over seventeen hundred. The wavy equals sign means approximately: about seventeen point six metres per second. So, twenty five on the level road, but only about seventeen point six on this gentle slope. Same engine, same power, and now you know exactly why. Clear?

7:41Quick revision: slope, full power, lift and pump power

So, quick revision. At a steady speed up a slope, P equals the bracket m g sine theta plus R, times v. At full power, the greatest speed is P over the total resisting force. For a lift, P equals M g plus f, times v; and for a pump, P equals m dot, times g h plus half v squared. Now go and try the practice problems on this page. You will do them well, I promise. All the best!