# Work: Example 1

by Jared Rovny
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00:01 Let's try to apply this to a problem in which we have maybe a 5 kg mass, initially at some velocity 4 meters per second, and then we apply force over a distance, pushing our object and speeding it up, giving it more kinetic energy.

00:13 We do this over a distance of 2 meters and we want to know first, what is the final kinetic energy of the object, and then, how much work was done on our object.

00:22 So, use the expressions that we've just found for work as a force times a distance in an angle between these 2 and see if you can find both of these quantities, the change in kinetic energy, and the work that was done.

00:33 If you did this, hopefully, it'll look something like this.

00:37 You have an object that initially has some velocity, which means it has some initial kinetic energy.

00:41 We apply a force to our object over some distance, in this case, it's going to be a distance of 2 meters.

00:47 After we've done this, because we've applied a force, the velocity is going to be much greater.

00:52 We've pushed the object in this direction of motion and that picked up more and more velocity which gave it a new final kinetic energy which is greater than the initial kinetic energy.

01:00 So writing this all out, we can see that the total amount of work that was done is equal to the force times the distance times the cosine of the angle between these two.

01:14 Now, because the direction of motion is to the right, and because the force is also to the right, the angle between them is zero and so this cosine of theta, which is zero, is equal to 1.

01:26 So, now we just get force times distance for our work.

01:30 So, work is equal to the force times the distance, which means, from our work energy theorem, the total work done on the object equals the change, that's what delta means, the change of kinetic energy, which is the final kinetic energy minus the initial kinetic energy.

01:47 From this we can say, and let's actually not evaluate this any further just yet, let's say that the final kinetic energy is what we're looking for first.

02:00 So, we say the final kinetic energy equals the work done plus the initial kinetic energy.

02:05 And this is a very key way of thinking about work.

02:08 Your final kinetic energy will be whatever energy you started with plus however much it was changed by work.

02:14 Evaluating this expression using the value for work that we just found of force times distance, we have the final kinetic energy is force times distance plus the initial 1/2 mv squared.

02:28 So now we can put in some numbers here and find the final kinetic energy which is the force times distance, which is 10 Newtons times 2 meters.

02:38 We're going to add that to the initial kinetic energy which was 1/2 our mass which is 5 times our velocity which is 4 squared, so this equals 20 plus 4 squared is 16, half of that is 8, so 8 times 5 is 40, so this equals 60.

02:59 And what are the units here, we have a kinetic energy and everything we've used as in our standard units.

03:04 We have kilograms, we have meters per second, we have Newton.

03:08 So everything is in our standard units which means we don't have to worry about any unit conversions and so we get our standard units and what we expect is an answer in terms of joules.

03:17 The second part of this question asks, how much work was done by this force? We actually already calculated it.

03:25 The work was the force times the distance, which was this right here.

03:28 The force times the distance which is coming from work.

03:32 So this force times the distance as we saw here, is 20 joules of work.

03:37 So, by adding 20 joules of work to our object using the work being in the same direction, we've added 20 joules of energy rather to the object's kinetic energy and it has far more kinetic energy than it originally had.

03:51 Let's look in a little more detail now about the importance of direction when considering work and how to find this cosine of theta term.

04:01 The key scenarios that you should really be very comfortable with and familiar with are these: First of all, let's take a slope example as a good way of thinking about all these different possible directions and forces.

04:13 One kind of force might be the force up the slope fighting the motion of the object down the slope.

04:19 So in this case, we have friction as a good example of something that doesn't want you moving down the slope, so we have a force opposite your direction of motion.

04:25 If you have a force up the slope while your motion is down the slope, the direction between these two as you can see in the red and blue arrows here is 180 degrees.

04:36 They're pointing in opposite directions.

04:37 And so in this case, the cosine of 180 degrees will be a minus 1, meaning that your work will be negative, meaning that the friction is trying to slow your object down.

04:46 On the other hand, we could consider the horizontal or down the slope rather aspect of or component of the gravitational force which we saw was Fg times the sine of theta.

04:57 So this force is acting down the slope which is in the same direction as the motion of your object, the same direction as its velocity.

05:04 And when these are in the same direction, as you can see with the arrows here, the angle between them is now zero degrees.

05:10 In other words, they're pointing in the same direction and when that's the case, the work is positive.

05:15 It's trying to push your object along and give it more kinetic energy.

05:17 Finally, we might have some perpendicular forces, so in this example, we have forces into the slope like the component of the gravitational force which is into the slope, Fg cosine of theta, and we also could have forces directly out of the slope as we discussed the normal force which keeps the object from going into the slope.

05:36 But these 2 forces both into or out of the slope in what we've often called the y direction are both perpendicular to the direction of motion which is down the slope. And so, when you analyze the cosine of theta, where theta, the angle in this case, is a perfect 90 degrees for right angles, you get a cosine of 90 is zero.

05:53 In other words, these forces into and out of the slope are doing no work on the object because they're not pushing it to gain kinetic energy, they're not pulling it to stop its kinetic energy, there's no motion in the direction of these forces and for that reason, there's no work done by these forces.

06:06 So, in an example like this one, if you're trying to find the total work done by all of these forces, you would want to add up the work done by friction and the work done by the component of gravity that's going down the slope, and then you could find the change and the kinetic energy of your object as it moves down the slope.

The lecture Work: Example 1 by Jared Rovny is from the course Work.

1. 4000J
2. 6000J
3. 3000J
4. 2500J
5. 3200J
1. 2m
2. 3m
3. 1km
4. 2.3m
5. 2.5m

### Customer reviews

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Very simple explanation. Intuitive in nature.
By Aaishah F. on 25. November 2017 for Work: Example 1

Very simply explained. He made a concept that I had been struggling with seem very simple. He explained better than my physics instructor who seems to have given up on teaching a subject that is very interesting.