Work done by forces (part 1)

0:00- [Instructor] A block of mass m slides with an initial speed of v0 at t equals zero on a horizontal surface. The coefficient of kinetic friction between the block and the surface is mu k. How far does the block slide before coming to rest? Express the distance in terms of m, v0, mu k, and physical constants as appropriate.

0:19Okay, so what's given to us is that we have a block of mass m that already has a speed of v0 at time t equals zero. It's sliding on a rough horizontal surface. There's some force of friction over there. The question is, how far does the block slide before coming to a rest?

0:32I mean, because there's a force of friction, we would expect the block to slow down and then eventually come to a stop, right? So the question is, how far does it slide before coming to a stop? How do we do this? Well, one way is we can use the free body diagrams to calculate the net force on this block, and then from there, use the second law to calculate the acceleration as a function of time.

0:54And then from the acceleration, we can calculate velocity as a function of time. And then from there, we can calculate the position as a function of time, and then we can find out how much the position has changed. So that's one way to do that, but there is a much quicker way to get to the answer over here, and that's what we're gonna explore in this video.

1:10And that is by using the idea of work. To quickly recall the idea of work, suppose there is a block with a force F that acts on it, and it gets displaced by d while that force is acting. Then we define the work done by this force as the dot product of the force and the displacement.

1:31And just to recap dot products, the dot product between any two vectors is the product of their magnitudes and the cosine of the angle between them. So if the angle between the force and displacement is theta, then this dot product basically equals the magnitude of force times the magnitude of displacement times cosine of the angle between them.

1:50That's how you calculate work. Well, the first thing we can see about work is that it's a scalar quantity. Dot products always give you a scalar quantity, so it's just a number. Okay, but geometrically, what does F d cosine theta represent? Well, remember, F cosine theta represents the component of the force in the direction of the displacement.

2:13So when we are calculating work, what we are doing is we are calculating the product of the force parallel to the displacement and the displacement. That's what we are multiplying; that's how we calculate the work done. Okay, so since it's a scalar and it's just a number, it can be a positive number, it can be zero, or it can be a negative number.

2:35So let's consider the different cases just to get a feeling for all of this. Suppose the angle theta was zero, then notice the entire force is in the direction of the displacement, right? And so now the work done will be just F times d. And that's exactly what this equation is saying.

2:51Cosine of zero is one. And so the work done is F times d. And so you get the maximum value of work done, you get a positive number, and you get the biggest positive number. Okay. What happens as the angle increases? Well, as the angle increases, the component in the direction of the displacement, the parallel component, that starts reducing.

3:11And so the work done becomes smaller and smaller, but as long as theta is acute, cosine is positive. And so the work done remains positive. What that means is, as long as theta is acute, the component of the force is in the direction of the displacement. And so the work done is a positive number.

3:30But what happens when theta becomes 90 degrees? Well, now cosine 90 is zero. So the work done becomes zero. Does that make sense? Yeah. Because now there is no component of the force in the direction of displacement, right? So there is no parallel component, and so there is no work done.

3:48Okay. What if theta was obtuse? Well, then notice cosine of an obtuse angle is a negative number, and so the work done now becomes negative. And again, what does that mean geometrically? Well, now, when you calculate the component of the force in the direction of displacement, it's in the opposite direction.

4:07That's why the work done is negative, because that component, the parallel component, is in the opposite direction. And now, as the angle becomes larger and larger, and eventually the angle becomes 180 degrees, notice cosine of 180 degrees is minus one. So now the work done becomes negative F d.

4:27Now the entire force is in the opposite direction, so you get negative F times d. So the angle between the force and displacement plays a crucial role in calculating the work done. But what if there are more than one forces acting on an object? Well, then we can calculate the work done due to individual forces this way.

4:45So for example, if this force was F1, then this would be the work done by the force F1. And now let's say there's another force F2 acting on it, making an angle of theta two with the displacement, then the work done by that force will be calculated in a similar way.

5:00So you can calculate the work done by each force on that object, and then you can calculate the total work done by just adding them up. And that's one of the things I love about work done because, remember, work is a scalar quantity. So we don't have to worry about, you know, directions and stuff.

5:15It's just a number. So the total work done on this object would be the sum of the work done by each force. But at this point, you may be wondering, well, okay, but what does this have anything to do with this problem? Right? Why are we talking about work done and stuff?

5:31Well, because there is a powerful principle, the change in the kinetic energy of an object always equals the total work done on that object. We call this the work-energy theorem. And notice kinetic energy has speed in it. So there's v is over here, and work done has displacement terms in it.

5:51And so we can use this equation to connect displacement and speeds to arrive at an expression. And this is especially useful when we are not asked to calculate, you know, speeds or positions as a function of time, because there is no time in this expression over here.

6:08Right? And look, in our question, we are not asked how long it takes. We are just asked how much, you know, it gets displaced before coming to a stop. So in these kinds of problems, using the work-energy theorem can give us the result much faster. So end of recap, let's now use this to solve our problem.

6:26So what's the change in the kinetic energy? Well, change in kinetic energy is the final kinetic energy minus the initial kinetic energy of the block. And that should equal the total work done on this block. What's the final kinetic energy? Well, that's the kinetic energy when the block is over here, when the block has come to a stop, in other words, when the velocity is zero.

6:46Since kinetic energy is half mv squared, the final kinetic energy is zero. Okay, minus what's the initial kinetic energy? Our initial point is over here at time t equals zero. And we know that the block already has a speed of v0. So the initial kinetic energy is half m v0 squared, and that should equal the total work done on the block.

7:08So now we need to figure out what's the total work done on the block. And we do that by adding the work done by all the forces on this block. So the first thing is we need to figure out all the forces acting on the block. And for that, let's draw a free body diagram.

7:23Great idea to pause the video and see if you can try drawing the free body diagram yourself. Okay, so what are the different forces acting on it? Well, there is the force of gravity, that is mg, then the surface will push up on it. So that's the normal force. And then there's the force of friction.

7:40The force of friction opposes relative motion. In this case, it opposes sliding motion. So the block is sliding to the right, and so it opposes it. So the force will be to the left. So that's the force of friction. And you might recall the force of friction while sliding is the kinetic, you know, coefficient of kinetic friction times the normal force.

8:00So we know the force of gravity, we don't know the normal force, we don't know the force of friction. In fact, if we figure out the normal force, we get the force of friction. So we just need to figure out the normal force. How do we do that? Well, we can apply Newton's second law in the vertical.

8:12We can do this in our heads. The block is not accelerating up or down. So the acceleration in the vertical is zero. So the net force in the vertical must be zero. In other words, these two forces must cancel each other. They must be balanced. So the normal force should equal the force of gravity, that should equal mg.

8:28And once I have the normal force, I can plug that in over here. So our force of friction becomes mu k times mg. Now that I have all the three forces, I can calculate the work done by all of these forces on the block and add them up to get the total work done. Now I can definitely do that, but look, these two forces are equal and opposite.

8:46So their contribution to work will just cancel out. And by the way, these two forces don't do work anyways because, look, their direction is perpendicular to the direction of the displacement. So since you're finding the net work done by using the net force, these two forces balance out.

9:03And so I can completely ignore them. So the only force that contributes to total work done is the force of friction. So we know the force of friction, that is mu k mg. And what is the direction of displacement? We know the displacement is to the right, and that is d.

9:18This is not a force, okay? This is the displacement. Okay? So now again, if you hadn't paused before, it'll be a great idea to pause and now calculate what the total work done would be. That'll be just the work done by the friction. Plug it over here and then calculate the displacement.

9:34All right, so let's do this. Let's clean this up a little bit. So there we go. So the total work done is basically the work done by the force of friction. The work done by the force of friction would be the force of friction, F, times d, times cosine theta. So the force of friction, whose magnitude is mu k mg, times the displacement, whose magnitude is d, times cosine of the angle between these two.

9:54Well, that'd be cosine of 180 degrees. So if you simplify the left-hand side, you get minus half m v0 squared. Cosine of 180 degrees is just minus one. So you get negative mu k mg times d, and the mass cancels out. And look, I have the expression. If I rearrange, d equals v0 squared by two mu k times g.

10:15There we have it. That's how much it travels before coming to a stop. And look, it just took us a couple of steps to arrive at the answer. That's how powerful the work-energy theorem can be. On the other hand, if the question was about time, for example, how long it takes for this block to come to a stop, well, then we can't use the work-energy theorem anymore because there is no time over here.

10:39Now we'll have to go back to Newton's second law. Anyways, before winding up, whenever we derive an expression, it's a good idea to do a sense check. For example, here we see that the displacement is directly related to the initial speed. Does that make sense?

10:53Well, yeah, if the initial speed was higher, we would expect it to travel farther before coming to a stop. So we would expect the displacement d to be larger. On the other hand, d is inversely related to mu k. Does that make sense? Yeah. If mu k was larger, the force of friction would be larger.

11:10If the force of friction is larger, imagine this is a much rougher surface, then you would expect it to come to a stop quicker, much closer. So you would expect d to be smaller. So it makes sense that it's in the denominator. What about the g in the denominator?

11:25Well, if the g was larger, then the force of gravity would be larger. The block would be pressed more against the surface. And again, for the given, you know, coefficient of kinetic friction, the force of friction would be larger. And again, d would be smaller.

11:40So it makes sense that it's in the denominator. Okay. Another sense check we can do is look at the units. The unit on the left-hand side is just like, you know, distance, so that's meters. So the unit on the right-hand side should also be meters. Let's just confirm that.

11:55On the numerator, you have speed squared; the unit of speed is just meters per second. So when you square it, you'll get meters squared per second squared. In the denominator, we have two, which is unitless. Mu k, what are its units? Well, mu k is a ratio of forces.

12:10Remember, this is just the normal force. So it's just a ratio of two forces. So mu k is unitless. The only thing that has units over here would be acceleration due to gravity, which is meters per second squared. And so the second squared cancels, and the meter and the meter cancels.

12:25Perfect. We are left with just meters. Yeah, it all checks out.