Work Done by a Constant Force

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Picture a porter at a railway station, standing still for ten minutes with a thirty kilogram load on his head. His neck aches, and he is sweating. So how much work is he doing on that load? Most of you will say, a lot, right? Hold on to that answer. By the end of this lesson, you will see what physics says. This idea starts the whole unit on work, energy and power, and JEE questions on it come up often. So, let's begin.

0:33Work: only the force along the motion counts

Look at the picture. Here, a block of mass m is pulled by a constant force F, the orange arrow, at an angle theta, a Greek letter, above the floor. The little arrow on top means a vector. The block moves through a displacement s, also a vector, along the floor. The force has two parts: F cos theta along the floor, and F sin theta straight up. Does the upward part help move the block? No, not at all! The block does not move upward. So only F cos theta, the part along the displacement, does work. Got it? So, work is this component times the displacement. W equals F vector dot s vector, which equals F s cos theta. The dot is the dot product, and theta is the angle between the two vectors, okay? This one line is the heart of today's lesson. This is for a constant force; variable forces come later in this chapter. And work is a scalar: it has a size and a sign, but no direction.

1:34Positive, negative or zero work

Clear? Now, cos theta can be positive, zero or negative, and so can work. For theta from zero up to just below ninety degrees, W is greater than zero: the force helps the motion. The less-than sign with a line below means less than or equal to; the small circle means degrees. Now, what if theta is exactly ninety? Yes, exactly! Zero! Cos ninety degrees is zero, so a force perpendicular to the displacement does no work at all. With me? Above ninety, up to one hundred and eighty degrees, cos theta is negative. So W is less than zero: the force opposes the motion. Okay? One more case. If the point where the force acts does not move, there is no work. The double arrow on screen means implies. Remember our porter? His load does not move, so he does zero work on it. Puzzle solved! Clear?

2:32Units of work

Next, units. The S I unit is the joule, capital J, and one joule is one newton metre, N m. The dimensional formula, in square brackets, is M L squared T to the minus two. Got it? One erg is ten to the minus seven joule, and one electron volt, e V, is one point six times ten to the minus nineteen joule. Now, one kilowatt hour, k W h. How many joules? One thousand watts, capital W for watt here, times three thousand six hundred seconds: three point six times ten to the six joules. Very good!

3:08Work done by gravity and friction

Gravity first: W g, W with a small g below, equals minus m g delta h. Delta, the little triangle, means change, so delta h is the change in height. That is m g times h i minus h f, initial minus final height, whatever the path. Positive when the body comes down. Okay? Kinetic friction on a fixed surface: W f equals minus mu k N s. Mu k, Greek mu with a small k, is the coefficient of kinetic friction, capital N the normal reaction, and s the distance slid. Always negative. Right? In the picture, a block is pulled up a rough incline: gravity m g down, N perpendicular to the slope, the pull F up the slope, friction, small f, down it, and the dashed arrow s, the displacement, up the slope. Here W f equals minus mu m g cos theta times s, with theta the angle of the incline. Clear? For several constant forces, just add: W net equals sigma W i, where the big Greek sigma means sum. With me?

4:15Forces that do no work, and a trap

Now, some forces that do no work at all. First, the normal reaction on a body sliding on a fixed surface: it is perpendicular to the motion. Okay? Next, the tension in a pendulum string, and any centripetal force. The upside-down T on screen means perpendicular: both are perpendicular to the velocity. Right? Two more: the magnetic force on a moving charge, and friction on a wheel rolling without slipping on fixed ground, because its point of application does not move. So, the normal reaction never does work. True? No, no, careful! This is the trap. On a moving surface, like the floor of a lift or the bed of an accelerating truck, the normal reaction and static friction can do work on the body resting there. Always ask: does the point where the force acts move? Got it?

5:07Example 1: work by a rope pulling at 60 degrees

First example. Here, a five kilogram box, k g on screen, is pulled ten metres along a level floor by a rope pulling with forty newtons, at sixty degrees above the horizontal. Find the work done by the rope. Pause the video, and try it first. Done? Okay, let's check. The force is constant, so W equals F s cos theta, with theta the sixty degrees between the rope and the direction of motion. Forty times ten times cos sixty degrees. Okay? Cos sixty degrees is one half, so forty times ten times one half: two hundred joules. But wait. We never used the five kilograms. Is something wrong? Nothing is wrong! The rope's work needs only its force, the displacement and the angle. The mass is not needed; questions sometimes give extra data. Clear?

6:01Example 2: work by each force on a rough incline

Now a harder one. Here, a ten kilogram block is pulled five metres up a rough incline of thirty seven degrees; mu, the coefficient of friction, is zero point five. The pull is parallel to the incline, at constant speed. Find the work done by each force. Sin thirty seven degrees is zero point six, and g is ten metres per second squared, written m s to the minus two. Pause and try. Hint: list all four forces. Done? Okay, let's check. Gravity first, with theta now thirty seven degrees. Its part along the slope, m g sin theta, points down, but the block moves up. So W g is minus ten times ten times zero point six times five: minus three hundred joules. Right? Now friction. For that we need cos thirty seven degrees, and it is not given. What is it? Zero point eight, since sin squared plus cos squared is one. Very good! Friction acts down the slope, and m g is one hundred newtons. So W f is minus zero point five times one hundred times zero point eight times five: minus two hundred joules. Okay? The normal reaction is perpendicular to the displacement, so W N, its work, is zero. Now the pull. Its size is not given! How do we find it? Constant speed! Yes, exactly! So the forces along the slope balance. The pull equals gravity's part, sixty newtons, plus friction, forty newtons: one hundred newtons. With me? The pull is along the motion, so its work is positive: one hundred times five, plus five hundred joules. Got it? Add them, with signs. Five hundred minus three hundred minus two hundred plus zero: zero! At constant speed the net force is zero, so the net work must be zero. That's it!

7:52Quick revision: W = Fs cos theta and the sign of work

So, quick revision. W equals F s cos theta: only the part of the force along the displacement does work. The angle decides the sign: below ninety positive, at ninety zero, above ninety negative. No displacement of the point of application, no work, like our porter. Now try the practice problems on this page. All the best!